The frequency of the wave.

Question

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Answer 1

Question

Chapter 16, Problem 16.13P

(a)

To determine

The frequency of the wave.

(a)

Expert Solution

Answer to Problem 16.13P

The frequency of the wave is $0.500 Hz$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula to calculate frequency of wave is,

$f=vλ$

Here,

$f$ is frequency of wave.

$v$ is speed of wave.

$λ$ is wavelength of wave.

Substitute $1.00 m/s$ for $v$ and $2.00 m$ for $λ$ in the above expression.

$f=1.00 m/s2.00 m=0.500 Hz$

Conclusion:

Therefore, the frequency of the wave is $0.500 Hz$ .

(b)

To determine

The angular frequency of the wave.

(b)

Expert Solution

Answer to Problem 16.13P

The angular frequency of the wave is $3.14 rad/s$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula to calculate angular frequency of the wave is,

$ω=2πf$

Here,

$ω$ is angular frequency of the wave.

Substitute $0.5 Hz$ for $f$ in the above expression.

$ω=2π(0.5 Hz)=π rad/s=3.14 rad/s$ (1)

Conclusion:

Therefore, the angular frequency of the wave is $3.14 rad/s$ .

(c)

To determine

The angular wave number of the wave.

(c)

Expert Solution

Answer to Problem 16.13P

The angular wave number of the wave is $3.14 rad/m$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula to calculate angular wave number of the wave is,

$k=2πλ$

Here,

$k$ is angular wave number.

Substitute $2.00 m$ for $λ$ in the above expression.

$k=2π2.00 m=π rad/m=3.14 rad/m$ (2)

Conclusion:

Therefore, the angular wave number of the wave is $3.14 rad/m$ .

(d)

To determine

The wave function of the wave.

(d)

Expert Solution

Answer to Problem 16.13P

The wave function of the wave is $0.100sin(πx−πt)$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula of standard wave equation is,

$y=Asin(kx−ωt)$

Here,

$A$ is amplitude of the wave.

$k$ is angular wave number of the wave.

$t$ is time period of wave.

Substitute $0.100 m$ for $A$ , $π rad/m$ for $k$ and $π rad/s$ for $ω$ in the above equation.

$y=(0.100 m)sin(πx−πt)$ (3)

Conclusion:

Therefore, the function of the wave is $0.100sin(πx−πt)$ .

(e)

To determine

The equation of motion for the left end of string.

(e)

Expert Solution

Answer to Problem 16.13P

The equation of motion for the left end of string is $0.100sin(−πt)$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

From equation (3),

$y=(0.100 m)sin(πx−πt)$

For the left end of string the position coordinate $x$ of the wave is $0$ .

Substitute $0$ for $x$ in the above equation.

$y=(0.100 m)sin(π(0)−πt)=(0.100 m)sin(−πt)$

Conclusion:

Therefore, the equation of motion for the left end of string is $0.100sin(−πt)$ .

(f)

To determine

The point on the string $x=1.50 m$ to the right of left end.

(f)

Expert Solution

Answer to Problem 16.13P

The equation of motion for the left end of string is $0.100sin(4.71−πt)$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

From equation (3),

$y=(0.100 m)sin(πx−πt)$

For the point $1.50 m$ to the right of the left end of the string:

Substitute $1.50 m$ for $x$ in the above expression.

$y=(0.100 m)sin(π(1.50 m)−πt)$

Solve the above expression for $y$ ,

$y=(0.100 m)sin(4.71−πt)$

Conclusion:

Therefore, the equation of motion for the left end of string is $0.100sin(4.71−πt)$ .

(g)

To determine

The maximum speed of element of the string.

(g)

Expert Solution

Answer to Problem 16.13P

The maximum speed of element of string is $0.314 m/s$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

From equation (3), the position of the wave is,

$y=(0.100 m)sin(πx−πt)$

The change in position with respect to time gives the speed.

Differentiate above equation n with respect to time,

$v=dydt$

Here,

$v$ is speed of the element of string.

Substitute $(0.100 m)sin(πx−πt)$ for $y$ in the above expression.

$v=d((0.100 m)sin(πx−πt))dt$

Solve the above expression for $v$ ,

$v=d((0.100 m)sin(πx−πt))dt=(0.100 m)dsin(πx−πt)dt+sin(πx−πt)d(0.100 m)dt=(0.100 m)cos(πx−πt)(−π)+0$

Substitute $3.14$ for $π$ in the above expression.

$=(0.100 m)cos((3.14)x−(3.14)t)(−3.14)+0=−0.314 mcos((3.14)x−(3.14)t)$

As the cosine wave varies from the positive of $+1$ to the negative of $+1$ . So the maximum speed of any element of string is $0.314 m/s$ .

Conclusion:

Therefore, the maximum speed of element of string is $0.314 m/s$ .

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Answer 2

Question

Chapter 16, Problem 16.13P

(a)

To determine

The frequency of the wave.

(a)

Expert Solution

Answer to Problem 16.13P

The frequency of the wave is $0.500 Hz$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula to calculate frequency of wave is,

$f=vλ$

Here,

$f$ is frequency of wave.

$v$ is speed of wave.

$λ$ is wavelength of wave.

Substitute $1.00 m/s$ for $v$ and $2.00 m$ for $λ$ in the above expression.

$f=1.00 m/s2.00 m=0.500 Hz$

Conclusion:

Therefore, the frequency of the wave is $0.500 Hz$ .

(b)

To determine

The angular frequency of the wave.

(b)

Expert Solution

Answer to Problem 16.13P

The angular frequency of the wave is $3.14 rad/s$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula to calculate angular frequency of the wave is,

$ω=2πf$

Here,

$ω$ is angular frequency of the wave.

Substitute $0.5 Hz$ for $f$ in the above expression.

$ω=2π(0.5 Hz)=π rad/s=3.14 rad/s$ (1)

Conclusion:

Therefore, the angular frequency of the wave is $3.14 rad/s$ .

(c)

To determine

The angular wave number of the wave.

(c)

Expert Solution

Answer to Problem 16.13P

The angular wave number of the wave is $3.14 rad/m$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula to calculate angular wave number of the wave is,

$k=2πλ$

Here,

$k$ is angular wave number.

Substitute $2.00 m$ for $λ$ in the above expression.

$k=2π2.00 m=π rad/m=3.14 rad/m$ (2)

Conclusion:

Therefore, the angular wave number of the wave is $3.14 rad/m$ .

(d)

To determine

The wave function of the wave.

(d)

Expert Solution

Answer to Problem 16.13P

The wave function of the wave is $0.100sin(πx−πt)$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula of standard wave equation is,

$y=Asin(kx−ωt)$

Here,

$A$ is amplitude of the wave.

$k$ is angular wave number of the wave.

$t$ is time period of wave.

Substitute $0.100 m$ for $A$ , $π rad/m$ for $k$ and $π rad/s$ for $ω$ in the above equation.

$y=(0.100 m)sin(πx−πt)$ (3)

Conclusion:

Therefore, the function of the wave is $0.100sin(πx−πt)$ .

(e)

To determine

The equation of motion for the left end of string.

(e)

Expert Solution

Answer to Problem 16.13P

The equation of motion for the left end of string is $0.100sin(−πt)$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

From equation (3),

$y=(0.100 m)sin(πx−πt)$

For the left end of string the position coordinate $x$ of the wave is $0$ .

Substitute $0$ for $x$ in the above equation.

$y=(0.100 m)sin(π(0)−πt)=(0.100 m)sin(−πt)$

Conclusion:

Therefore, the equation of motion for the left end of string is $0.100sin(−πt)$ .

(f)

To determine

The point on the string $x=1.50 m$ to the right of left end.

(f)

Expert Solution

Answer to Problem 16.13P

The equation of motion for the left end of string is $0.100sin(4.71−πt)$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

From equation (3),

$y=(0.100 m)sin(πx−πt)$

For the point $1.50 m$ to the right of the left end of the string:

Substitute $1.50 m$ for $x$ in the above expression.

$y=(0.100 m)sin(π(1.50 m)−πt)$

Solve the above expression for $y$ ,

$y=(0.100 m)sin(4.71−πt)$

Conclusion:

Therefore, the equation of motion for the left end of string is $0.100sin(4.71−πt)$ .

(g)

To determine

The maximum speed of element of the string.

(g)

Expert Solution

Answer to Problem 16.13P

The maximum speed of element of string is $0.314 m/s$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

From equation (3), the position of the wave is,

$y=(0.100 m)sin(πx−πt)$

The change in position with respect to time gives the speed.

Differentiate above equation n with respect to time,

$v=dydt$

Here,

$v$ is speed of the element of string.

Substitute $(0.100 m)sin(πx−πt)$ for $y$ in the above expression.

$v=d((0.100 m)sin(πx−πt))dt$

Solve the above expression for $v$ ,

$v=d((0.100 m)sin(πx−πt))dt=(0.100 m)dsin(πx−πt)dt+sin(πx−πt)d(0.100 m)dt=(0.100 m)cos(πx−πt)(−π)+0$

Substitute $3.14$ for $π$ in the above expression.

$=(0.100 m)cos((3.14)x−(3.14)t)(−3.14)+0=−0.314 mcos((3.14)x−(3.14)t)$

As the cosine wave varies from the positive of $+1$ to the negative of $+1$ . So the maximum speed of any element of string is $0.314 m/s$ .

Conclusion:

Therefore, the maximum speed of element of string is $0.314 m/s$ .

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Answer 3

Question

Chapter 16, Problem 16.13P

(a)

To determine

The frequency of the wave.

(a)

Expert Solution

Answer to Problem 16.13P

The frequency of the wave is $0.500 Hz$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula to calculate frequency of wave is,

$f=vλ$

Here,

$f$ is frequency of wave.

$v$ is speed of wave.

$λ$ is wavelength of wave.

Substitute $1.00 m/s$ for $v$ and $2.00 m$ for $λ$ in the above expression.

$f=1.00 m/s2.00 m=0.500 Hz$

Conclusion:

Therefore, the frequency of the wave is $0.500 Hz$ .

(b)

To determine

The angular frequency of the wave.

(b)

Expert Solution

Answer to Problem 16.13P

The angular frequency of the wave is $3.14 rad/s$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula to calculate angular frequency of the wave is,

$ω=2πf$

Here,

$ω$ is angular frequency of the wave.

Substitute $0.5 Hz$ for $f$ in the above expression.

$ω=2π(0.5 Hz)=π rad/s=3.14 rad/s$ (1)

Conclusion:

Therefore, the angular frequency of the wave is $3.14 rad/s$ .

(c)

To determine

The angular wave number of the wave.

(c)

Expert Solution

Answer to Problem 16.13P

The angular wave number of the wave is $3.14 rad/m$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula to calculate angular wave number of the wave is,

$k=2πλ$

Here,

$k$ is angular wave number.

Substitute $2.00 m$ for $λ$ in the above expression.

$k=2π2.00 m=π rad/m=3.14 rad/m$ (2)

Conclusion:

Therefore, the angular wave number of the wave is $3.14 rad/m$ .

(d)

To determine

The wave function of the wave.

(d)

Expert Solution

Answer to Problem 16.13P

The wave function of the wave is $0.100sin(πx−πt)$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula of standard wave equation is,

$y=Asin(kx−ωt)$

Here,

$A$ is amplitude of the wave.

$k$ is angular wave number of the wave.

$t$ is time period of wave.

Substitute $0.100 m$ for $A$ , $π rad/m$ for $k$ and $π rad/s$ for $ω$ in the above equation.

$y=(0.100 m)sin(πx−πt)$ (3)

Conclusion:

Therefore, the function of the wave is $0.100sin(πx−πt)$ .

(e)

To determine

The equation of motion for the left end of string.

(e)

Expert Solution

Answer to Problem 16.13P

The equation of motion for the left end of string is $0.100sin(−πt)$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

From equation (3),

$y=(0.100 m)sin(πx−πt)$

For the left end of string the position coordinate $x$ of the wave is $0$ .

Substitute $0$ for $x$ in the above equation.

$y=(0.100 m)sin(π(0)−πt)=(0.100 m)sin(−πt)$

Conclusion:

Therefore, the equation of motion for the left end of string is $0.100sin(−πt)$ .

(f)

To determine

The point on the string $x=1.50 m$ to the right of left end.

(f)

Expert Solution

Answer to Problem 16.13P

The equation of motion for the left end of string is $0.100sin(4.71−πt)$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

From equation (3),

$y=(0.100 m)sin(πx−πt)$

For the point $1.50 m$ to the right of the left end of the string:

Substitute $1.50 m$ for $x$ in the above expression.

$y=(0.100 m)sin(π(1.50 m)−πt)$

Solve the above expression for $y$ ,

$y=(0.100 m)sin(4.71−πt)$

Conclusion:

Therefore, the equation of motion for the left end of string is $0.100sin(4.71−πt)$ .

(g)

To determine

The maximum speed of element of the string.

(g)

Expert Solution

Answer to Problem 16.13P

The maximum speed of element of string is $0.314 m/s$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

From equation (3), the position of the wave is,

$y=(0.100 m)sin(πx−πt)$

The change in position with respect to time gives the speed.

Differentiate above equation n with respect to time,

$v=dydt$

Here,

$v$ is speed of the element of string.

Substitute $(0.100 m)sin(πx−πt)$ for $y$ in the above expression.

$v=d((0.100 m)sin(πx−πt))dt$

Solve the above expression for $v$ ,

$v=d((0.100 m)sin(πx−πt))dt=(0.100 m)dsin(πx−πt)dt+sin(πx−πt)d(0.100 m)dt=(0.100 m)cos(πx−πt)(−π)+0$

Substitute $3.14$ for $π$ in the above expression.

$=(0.100 m)cos((3.14)x−(3.14)t)(−3.14)+0=−0.314 mcos((3.14)x−(3.14)t)$

As the cosine wave varies from the positive of $+1$ to the negative of $+1$ . So the maximum speed of any element of string is $0.314 m/s$ .

Conclusion:

Therefore, the maximum speed of element of string is $0.314 m/s$ .

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Answer 4

Question

Chapter 16, Problem 16.13P

(a)

To determine

The frequency of the wave.

(a)

Expert Solution

Answer to Problem 16.13P

The frequency of the wave is $0.500 Hz$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula to calculate frequency of wave is,

$f=vλ$

Here,

$f$ is frequency of wave.

$v$ is speed of wave.

$λ$ is wavelength of wave.

Substitute $1.00 m/s$ for $v$ and $2.00 m$ for $λ$ in the above expression.

$f=1.00 m/s2.00 m=0.500 Hz$

Conclusion:

Therefore, the frequency of the wave is $0.500 Hz$ .

(b)

To determine

The angular frequency of the wave.

(b)

Expert Solution

Answer to Problem 16.13P

The angular frequency of the wave is $3.14 rad/s$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula to calculate angular frequency of the wave is,

$ω=2πf$

Here,

$ω$ is angular frequency of the wave.

Substitute $0.5 Hz$ for $f$ in the above expression.

$ω=2π(0.5 Hz)=π rad/s=3.14 rad/s$ (1)

Conclusion:

Therefore, the angular frequency of the wave is $3.14 rad/s$ .

(c)

To determine

The angular wave number of the wave.

(c)

Expert Solution

Answer to Problem 16.13P

The angular wave number of the wave is $3.14 rad/m$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula to calculate angular wave number of the wave is,

$k=2πλ$

Here,

$k$ is angular wave number.

Substitute $2.00 m$ for $λ$ in the above expression.

$k=2π2.00 m=π rad/m=3.14 rad/m$ (2)

Conclusion:

Therefore, the angular wave number of the wave is $3.14 rad/m$ .

(d)

To determine

The wave function of the wave.

(d)

Expert Solution

Answer to Problem 16.13P

The wave function of the wave is $0.100sin(πx−πt)$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula of standard wave equation is,

$y=Asin(kx−ωt)$

Here,

$A$ is amplitude of the wave.

$k$ is angular wave number of the wave.

$t$ is time period of wave.

Substitute $0.100 m$ for $A$ , $π rad/m$ for $k$ and $π rad/s$ for $ω$ in the above equation.

$y=(0.100 m)sin(πx−πt)$ (3)

Conclusion:

Therefore, the function of the wave is $0.100sin(πx−πt)$ .

(e)

To determine

The equation of motion for the left end of string.

(e)

Expert Solution

Answer to Problem 16.13P

The equation of motion for the left end of string is $0.100sin(−πt)$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

From equation (3),

$y=(0.100 m)sin(πx−πt)$

For the left end of string the position coordinate $x$ of the wave is $0$ .

Substitute $0$ for $x$ in the above equation.

$y=(0.100 m)sin(π(0)−πt)=(0.100 m)sin(−πt)$

Conclusion:

Therefore, the equation of motion for the left end of string is $0.100sin(−πt)$ .

(f)

To determine

The point on the string $x=1.50 m$ to the right of left end.

(f)

Expert Solution

Answer to Problem 16.13P

The equation of motion for the left end of string is $0.100sin(4.71−πt)$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

From equation (3),

$y=(0.100 m)sin(πx−πt)$

For the point $1.50 m$ to the right of the left end of the string:

Substitute $1.50 m$ for $x$ in the above expression.

$y=(0.100 m)sin(π(1.50 m)−πt)$

Solve the above expression for $y$ ,

$y=(0.100 m)sin(4.71−πt)$

Conclusion:

Therefore, the equation of motion for the left end of string is $0.100sin(4.71−πt)$ .

(g)

To determine

The maximum speed of element of the string.

(g)

Expert Solution

Answer to Problem 16.13P

The maximum speed of element of string is $0.314 m/s$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

From equation (3), the position of the wave is,

$y=(0.100 m)sin(πx−πt)$

The change in position with respect to time gives the speed.

Differentiate above equation n with respect to time,

$v=dydt$

Here,

$v$ is speed of the element of string.

Substitute $(0.100 m)sin(πx−πt)$ for $y$ in the above expression.

$v=d((0.100 m)sin(πx−πt))dt$

Solve the above expression for $v$ ,

$v=d((0.100 m)sin(πx−πt))dt=(0.100 m)dsin(πx−πt)dt+sin(πx−πt)d(0.100 m)dt=(0.100 m)cos(πx−πt)(−π)+0$

Substitute $3.14$ for $π$ in the above expression.

$=(0.100 m)cos((3.14)x−(3.14)t)(−3.14)+0=−0.314 mcos((3.14)x−(3.14)t)$

As the cosine wave varies from the positive of $+1$ to the negative of $+1$ . So the maximum speed of any element of string is $0.314 m/s$ .

Conclusion:

Therefore, the maximum speed of element of string is $0.314 m/s$ .

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Answer 5

Chapter 16, Problem 16.13P

(a)

To determine

The frequency of the wave.

(a)

Expert Solution

Answer to Problem 16.13P

The frequency of the wave is $0.500 Hz$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula to calculate frequency of wave is,

$f=vλ$

Here,

$f$ is frequency of wave.

$v$ is speed of wave.

$λ$ is wavelength of wave.

Substitute $1.00 m/s$ for $v$ and $2.00 m$ for $λ$ in the above expression.

$f=1.00 m/s2.00 m=0.500 Hz$

Conclusion:

Therefore, the frequency of the wave is $0.500 Hz$ .

(b)

To determine

The angular frequency of the wave.

(b)

Expert Solution

Answer to Problem 16.13P

The angular frequency of the wave is $3.14 rad/s$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula to calculate angular frequency of the wave is,

$ω=2πf$

Here,

$ω$ is angular frequency of the wave.

Substitute $0.5 Hz$ for $f$ in the above expression.

$ω=2π(0.5 Hz)=π rad/s=3.14 rad/s$ (1)

Conclusion:

Therefore, the angular frequency of the wave is $3.14 rad/s$ .

(c)

To determine

The angular wave number of the wave.

(c)

Expert Solution

Answer to Problem 16.13P

The angular wave number of the wave is $3.14 rad/m$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula to calculate angular wave number of the wave is,

$k=2πλ$

Here,

$k$ is angular wave number.

Substitute $2.00 m$ for $λ$ in the above expression.

$k=2π2.00 m=π rad/m=3.14 rad/m$ (2)

Conclusion:

Therefore, the angular wave number of the wave is $3.14 rad/m$ .

(d)

To determine

The wave function of the wave.

(d)

Expert Solution

Answer to Problem 16.13P

The wave function of the wave is $0.100sin(πx−πt)$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula of standard wave equation is,

$y=Asin(kx−ωt)$

Here,

$A$ is amplitude of the wave.

$k$ is angular wave number of the wave.

$t$ is time period of wave.

Substitute $0.100 m$ for $A$ , $π rad/m$ for $k$ and $π rad/s$ for $ω$ in the above equation.

$y=(0.100 m)sin(πx−πt)$ (3)

Conclusion:

Therefore, the function of the wave is $0.100sin(πx−πt)$ .

(e)

To determine

The equation of motion for the left end of string.

(e)

Expert Solution

Answer to Problem 16.13P

The equation of motion for the left end of string is $0.100sin(−πt)$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

From equation (3),

$y=(0.100 m)sin(πx−πt)$

For the left end of string the position coordinate $x$ of the wave is $0$ .

Substitute $0$ for $x$ in the above equation.

$y=(0.100 m)sin(π(0)−πt)=(0.100 m)sin(−πt)$

Conclusion:

Therefore, the equation of motion for the left end of string is $0.100sin(−πt)$ .

(f)

To determine

The point on the string $x=1.50 m$ to the right of left end.

(f)

Expert Solution

Answer to Problem 16.13P

The equation of motion for the left end of string is $0.100sin(4.71−πt)$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

From equation (3),

$y=(0.100 m)sin(πx−πt)$

For the point $1.50 m$ to the right of the left end of the string:

Substitute $1.50 m$ for $x$ in the above expression.

$y=(0.100 m)sin(π(1.50 m)−πt)$

Solve the above expression for $y$ ,

$y=(0.100 m)sin(4.71−πt)$

Conclusion:

Therefore, the equation of motion for the left end of string is $0.100sin(4.71−πt)$ .

(g)

To determine

The maximum speed of element of the string.

(g)

Expert Solution

Answer to Problem 16.13P

The maximum speed of element of string is $0.314 m/s$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

From equation (3), the position of the wave is,

$y=(0.100 m)sin(πx−πt)$

The change in position with respect to time gives the speed.

Differentiate above equation n with respect to time,

$v=dydt$

Here,

$v$ is speed of the element of string.

Substitute $(0.100 m)sin(πx−πt)$ for $y$ in the above expression.

$v=d((0.100 m)sin(πx−πt))dt$

Solve the above expression for $v$ ,

$v=d((0.100 m)sin(πx−πt))dt=(0.100 m)dsin(πx−πt)dt+sin(πx−πt)d(0.100 m)dt=(0.100 m)cos(πx−πt)(−π)+0$

Substitute $3.14$ for $π$ in the above expression.

$=(0.100 m)cos((3.14)x−(3.14)t)(−3.14)+0=−0.314 mcos((3.14)x−(3.14)t)$

As the cosine wave varies from the positive of $+1$ to the negative of $+1$ . So the maximum speed of any element of string is $0.314 m/s$ .

Conclusion:

Therefore, the maximum speed of element of string is $0.314 m/s$ .

Answer 6

Chapter 16, Problem 16.13P

(a)

To determine

The frequency of the wave.

(a)

Expert Solution

Answer to Problem 16.13P

The frequency of the wave is $0.500 Hz$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula to calculate frequency of wave is,

$f=vλ$

Here,

$f$ is frequency of wave.

$v$ is speed of wave.

$λ$ is wavelength of wave.

Substitute $1.00 m/s$ for $v$ and $2.00 m$ for $λ$ in the above expression.

$f=1.00 m/s2.00 m=0.500 Hz$

Conclusion:

Therefore, the frequency of the wave is $0.500 Hz$ .

Answer 7

Chapter 16, Problem 16.13P

(a)

To determine

The frequency of the wave.

Answer 8

(a)

Expert Solution

Answer to Problem 16.13P

The frequency of the wave is $0.500 Hz$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula to calculate frequency of wave is,

$f=vλ$

Here,

$f$ is frequency of wave.

$v$ is speed of wave.

$λ$ is wavelength of wave.

Substitute $1.00 m/s$ for $v$ and $2.00 m$ for $λ$ in the above expression.

$f=1.00 m/s2.00 m=0.500 Hz$

Conclusion:

Therefore, the frequency of the wave is $0.500 Hz$ .

Answer 13

(b)

To determine

The angular frequency of the wave.

(b)

Expert Solution

Answer to Problem 16.13P

The angular frequency of the wave is $3.14 rad/s$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula to calculate angular frequency of the wave is,

$ω=2πf$

Here,

$ω$ is angular frequency of the wave.

Substitute $0.5 Hz$ for $f$ in the above expression.

$ω=2π(0.5 Hz)=π rad/s=3.14 rad/s$ (1)

Conclusion:

Therefore, the angular frequency of the wave is $3.14 rad/s$ .

Answer 14

(b)

To determine

The angular frequency of the wave.

Answer 15

(b)

Expert Solution

Answer to Problem 16.13P

The angular frequency of the wave is $3.14 rad/s$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula to calculate angular frequency of the wave is,

$ω=2πf$

Here,

$ω$ is angular frequency of the wave.

Substitute $0.5 Hz$ for $f$ in the above expression.

$ω=2π(0.5 Hz)=π rad/s=3.14 rad/s$ (1)

Conclusion:

Therefore, the angular frequency of the wave is $3.14 rad/s$ .

Answer 20

(c)

To determine

The angular wave number of the wave.

(c)

Expert Solution

Answer to Problem 16.13P

The angular wave number of the wave is $3.14 rad/m$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula to calculate angular wave number of the wave is,

$k=2πλ$

Here,

$k$ is angular wave number.

Substitute $2.00 m$ for $λ$ in the above expression.

$k=2π2.00 m=π rad/m=3.14 rad/m$ (2)

Conclusion:

Therefore, the angular wave number of the wave is $3.14 rad/m$ .

Answer 21

(c)

To determine

The angular wave number of the wave.

Answer 22

(c)

Expert Solution

Answer to Problem 16.13P

The angular wave number of the wave is $3.14 rad/m$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula to calculate angular wave number of the wave is,

$k=2πλ$

Here,

$k$ is angular wave number.

Substitute $2.00 m$ for $λ$ in the above expression.

$k=2π2.00 m=π rad/m=3.14 rad/m$ (2)

Conclusion:

Therefore, the angular wave number of the wave is $3.14 rad/m$ .

Answer 27

(d)

To determine

The wave function of the wave.

(d)

Expert Solution

Answer to Problem 16.13P

The wave function of the wave is $0.100sin(πx−πt)$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula of standard wave equation is,

$y=Asin(kx−ωt)$

Here,

$A$ is amplitude of the wave.

$k$ is angular wave number of the wave.

$t$ is time period of wave.

Substitute $0.100 m$ for $A$ , $π rad/m$ for $k$ and $π rad/s$ for $ω$ in the above equation.

$y=(0.100 m)sin(πx−πt)$ (3)

Conclusion:

Therefore, the function of the wave is $0.100sin(πx−πt)$ .

Answer 28

(d)

To determine

The wave function of the wave.

Answer 29

(d)

Expert Solution

Answer to Problem 16.13P

The wave function of the wave is $0.100sin(πx−πt)$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

The formula of standard wave equation is,

$y=Asin(kx−ωt)$

Here,

$A$ is amplitude of the wave.

$k$ is angular wave number of the wave.

$t$ is time period of wave.

Substitute $0.100 m$ for $A$ , $π rad/m$ for $k$ and $π rad/s$ for $ω$ in the above equation.

$y=(0.100 m)sin(πx−πt)$ (3)

Conclusion:

Therefore, the function of the wave is $0.100sin(πx−πt)$ .

Answer 34

(e)

To determine

The equation of motion for the left end of string.

(e)

Expert Solution

Answer to Problem 16.13P

The equation of motion for the left end of string is $0.100sin(−πt)$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

From equation (3),

$y=(0.100 m)sin(πx−πt)$

For the left end of string the position coordinate $x$ of the wave is $0$ .

Substitute $0$ for $x$ in the above equation.

$y=(0.100 m)sin(π(0)−πt)=(0.100 m)sin(−πt)$

Conclusion:

Therefore, the equation of motion for the left end of string is $0.100sin(−πt)$ .

Answer 35

(e)

To determine

The equation of motion for the left end of string.

Answer 36

(e)

Expert Solution

Answer to Problem 16.13P

The equation of motion for the left end of string is $0.100sin(−πt)$ .

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

From equation (3),

$y=(0.100 m)sin(πx−πt)$

For the left end of string the position coordinate $x$ of the wave is $0$ .

Substitute $0$ for $x$ in the above equation.

$y=(0.100 m)sin(π(0)−πt)=(0.100 m)sin(−πt)$

Conclusion:

Therefore, the equation of motion for the left end of string is $0.100sin(−πt)$ .

Answer 41

(f)

To determine

The point on the string $x=1.50 m$ to the right of left end.

(f)

Expert Solution

Answer to Problem 16.13P

The equation of motion for the left end of string is $0.100sin(4.71−πt)$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

From equation (3),

$y=(0.100 m)sin(πx−πt)$

For the point $1.50 m$ to the right of the left end of the string:

Substitute $1.50 m$ for $x$ in the above expression.

$y=(0.100 m)sin(π(1.50 m)−πt)$

Solve the above expression for $y$ ,

$y=(0.100 m)sin(4.71−πt)$

Conclusion:

Therefore, the equation of motion for the left end of string is $0.100sin(4.71−πt)$ .

Answer 42

(f)

To determine

The point on the string $x=1.50 m$ to the right of left end.

Answer 43

(f)

Expert Solution

Answer to Problem 16.13P

The equation of motion for the left end of string is $0.100sin(4.71−πt)$ .

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

From equation (3),

$y=(0.100 m)sin(πx−πt)$

For the point $1.50 m$ to the right of the left end of the string:

Substitute $1.50 m$ for $x$ in the above expression.

$y=(0.100 m)sin(π(1.50 m)−πt)$

Solve the above expression for $y$ ,

$y=(0.100 m)sin(4.71−πt)$

Conclusion:

Therefore, the equation of motion for the left end of string is $0.100sin(4.71−πt)$ .

Answer 48

(g)

To determine

The maximum speed of element of the string.

(g)

Expert Solution

Answer to Problem 16.13P

The maximum speed of element of string is $0.314 m/s$ .

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

From equation (3), the position of the wave is,

$y=(0.100 m)sin(πx−πt)$

The change in position with respect to time gives the speed.

Differentiate above equation n with respect to time,

$v=dydt$

Here,

$v$ is speed of the element of string.

Substitute $(0.100 m)sin(πx−πt)$ for $y$ in the above expression.

$v=d((0.100 m)sin(πx−πt))dt$

Solve the above expression for $v$ ,

$v=d((0.100 m)sin(πx−πt))dt=(0.100 m)dsin(πx−πt)dt+sin(πx−πt)d(0.100 m)dt=(0.100 m)cos(πx−πt)(−π)+0$

Substitute $3.14$ for $π$ in the above expression.

$=(0.100 m)cos((3.14)x−(3.14)t)(−3.14)+0=−0.314 mcos((3.14)x−(3.14)t)$

As the cosine wave varies from the positive of $+1$ to the negative of $+1$ . So the maximum speed of any element of string is $0.314 m/s$ .

Conclusion:

Therefore, the maximum speed of element of string is $0.314 m/s$ .

Answer 49

(g)

To determine

The maximum speed of element of the string.

Answer 50

(g)

Expert Solution

Answer to Problem 16.13P

The maximum speed of element of string is $0.314 m/s$ .

Answer 51

(g)

Expert Solution

Answer 52

(g)

Expert Solution

Answer 53

Expert Solution

Answer 54

Explanation of Solution

Given info: The wavelength of wave is $2.00 m$ , the amplitude is $0.100 m$ . The speed of wave in string is $1.00 m/s$ . The left end of wave is at origin at $t=0$ .

From equation (3), the position of the wave is,

$y=(0.100 m)sin(πx−πt)$

The change in position with respect to time gives the speed.

Differentiate above equation n with respect to time,

$v=dydt$

Here,

$v$ is speed of the element of string.

Substitute $(0.100 m)sin(πx−πt)$ for $y$ in the above expression.

$v=d((0.100 m)sin(πx−πt))dt$

Solve the above expression for $v$ ,

$v=d((0.100 m)sin(πx−πt))dt=(0.100 m)dsin(πx−πt)dt+sin(πx−πt)d(0.100 m)dt=(0.100 m)cos(πx−πt)(−π)+0$

Substitute $3.14$ for $π$ in the above expression.

$=(0.100 m)cos((3.14)x−(3.14)t)(−3.14)+0=−0.314 mcos((3.14)x−(3.14)t)$

As the cosine wave varies from the positive of $+1$ to the negative of $+1$ . So the maximum speed of any element of string is $0.314 m/s$ .

Conclusion:

Therefore, the maximum speed of element of string is $0.314 m/s$ .

Answer 55

Ch. 16 - Prob. 16.1QQ Ch. 16 - A sinusoidal wave of frequency f is traveling...Ch. 16 - The amplitude of a wave is doubled, with no other...Ch. 16 - Suppose you create a pulse by moving the free end...Ch. 16 - Which of the following, taken by itself, would be...Ch. 16 - If one end of a heavy rope is attached to one end...Ch. 16 - Prob. 16.2OQ Ch. 16 - Rank the waves represented by the following...Ch. 16 - By what factor would von have to multiply the...Ch. 16 - When all the strings on a guitar (Fig. OQ16.5) are...

The frequency of the wave.

Concept explainers

The frequency of the wave.

Answer to Problem 16.13P

Explanation of Solution

The angular frequency of the wave.

Answer to Problem 16.13P

Explanation of Solution

The angular wave number of the wave.

Answer to Problem 16.13P

Explanation of Solution

The wave function of the wave.

Answer to Problem 16.13P

Explanation of Solution

The equation of motion for the left end of string.

Answer to Problem 16.13P

Explanation of Solution

The point on the string $x=1.50 m$ to the right of left end.

Answer to Problem 16.13P

Explanation of Solution

The maximum speed of element of the string.

Answer to Problem 16.13P

Explanation of Solution

Want to see more full solutions like this?

Chapter 16 Solutions

The frequency of the wave.

Concept explainers

The frequency of the wave.

Answer to Problem 16.13P

Explanation of Solution

The angular frequency of the wave.

Answer to Problem 16.13P

Explanation of Solution

The angular wave number of the wave.

Answer to Problem 16.13P

Explanation of Solution

The wave function of the wave.

Answer to Problem 16.13P

Explanation of Solution

The equation of motion for the left end of string.

Answer to Problem 16.13P

Explanation of Solution

The point on the string x=1.50 m<m:math><m:mrow><m:mi>x</m:mi><m:mo>=</m:mo><m:mn>1.50</m:mn><m:mtext> </m:mtext><m:mtext>m</m:mtext></m:mrow> </m:math> to the right of left end.

Answer to Problem 16.13P

Explanation of Solution

The maximum speed of element of the string.

Answer to Problem 16.13P

Explanation of Solution

Want to see more full solutions like this?

Chapter 16 Solutions

The point on the string $x=1.50 m$ to the right of left end.