A transverse wave on a siring is described by the wave function y = 0.120 sin ( π 8 x + 4 π t ) where x and y are in meters and t is in seconds. Determine (a) the transverse speed and (b) the transverse acceleration at t = 0.200 s for an element of the string located at x = 1.60 m. What are (c) the wavelength, (d) the period, and (e) the speed of propagation of this wave?

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Answer 1

Textbook Question

Chapter 16, Problem 16.15P

A transverse wave on a siring is described by the wave function

$y = 0.120 sin ( π 8 x + 4 π t )$

where x and y are in meters and t is in seconds. Determine (a) the transverse speed and (b) the transverse acceleration at t = 0.200 s for an element of the string located at x = 1.60 m. What are (c) the wavelength, (d) the period, and (e) the speed of propagation of this wave?

(a)

Expert Solution

To determine

The transverse speed for an element located at $1.60 m$ .

Answer to Problem 16.15P

The transverse speed for an element located at $1.60 m$ is $−1.51 m/s$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The standard equation of the transverse wave is,

$y=Asin(kx−ωt)$ . (1)

Here,

$A$ is amplitude of wave.

$k$ is angular wave number of wave.

$ω$ is angular velocity of wave.

The wave function give is,

$y=0.120sin(π8x+4πt)$ . (2)

Compare the equation (1) and equation (2).

$A=0.120 mk=π8ω=4π rad/s$

The formula to calculate frequency is,

$f=ω2π$

Substitute $4π rad/s$ for $ω$ in the above expression.

$f=4π rad/s2π=2.00 Hz$

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

$v=dydt$

Here,

$dy$ is change in position.

$dt$ is change in time.

Substitute $0.120sin(π8x+4πt)$ for $y$ in the above expression.

$v=d(0.120sin(π8x+4πt))dt$

Differentiate and solve the above expression for $v$ ,

$=(0.120)dsin(π8x+4πt)+sin(π8x+4πt)d(0.120)dtdt=(0.120)cos(π8x+4πt)(4π)+0$ (1)

Substitute $0.200 s$ for $t$ and $1.6 m$ for $x$ in the above expression.

$=(0.120)cos(π8(1.6 m)+4π(0.200 s))(4π)+0=(0.120)(4π)cos(0.2π+0.8π)=−1.5072 m/s≈−1.51 m/s$

Conclusion:

Therefore, the transverse speed for an element located at $1.60 m$ is $−1.51 m/s$ .

(b)

Expert Solution

To determine

To write: The transverse acceleration of the wave.

Answer to Problem 16.15P

The transverse acceleration of the wave is $0$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ . The time of the wave is $0.200 s$ and the position of element is $1.60 m$ .

The change in velocity with respect to time gives the acceleration.

$a=dvdt$ (2)

Here,

$dv$ is change in velocity.

$dt$ is change in time.

From equation (1), the speed is,

$v=(0.120)cos(π8x+4πt)(4π)+0$

Substitute $(0.120)cos(π8x+4πt)(4π)+0$ for $v$ in equation (2).

$a=d(0.120)cos(π8x+4πt)(4π)+0dt$

Differentiate and solve the above expression for $a$ .

$a=d(0.120)cos(π8x+4πt)(4π)+0dt=(0.120)×(−sin((π8x+4πt))×(4π)2)+0$

Substitute $0.200 s$ for $t$ and $1.60 m$ for $x$ in the above expression.

$(0.120)×(−sin((π8(1.60 m)+4π(0.200 s)))×(4π)2)+0$

Solve the above expression.

$=(0.120)×(−sin((π8(1.60 m)+4π(0.200 s)))×(4π)2)+0=−(0.120)(4π)2(−sin(π))=0$

Conclusion:

Therefore, the transverse acceleration of the wave is $0$ .

(c)

Expert Solution

To determine

The wavelength of the wave.

Answer to Problem 16.15P

The wavelength of the wave is $16.0 m$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The formula to calculate wavelength of the wave is,

$λ=2πk$

Here,

$k$ is angular wave number of wave.

Substitute $π8$ for $k$ in the above expression.

$λ=2π(π8)$

Solve the above expression for $λ$ .

$λ=2π(π8)=2×8 m=16.0 m$

Conclusion:

Therefore, the wavelength of the wave is $16.0 m$ .

(d)

Expert Solution

To determine

The period of the wave.

Answer to Problem 16.15P

The period of the wave is $0.500 s$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The formula to calculate frequency is,

$f=ω2π$

Here,

$ω$ is angular frequency of wave.

Substitute $4π rad/s$ for $ω$ in the above expression.

$f=4π rad/s2π=2 Hz$

The formula to calculate time period of the wave is,

$t=1f$

Here,

$f$ is frequency of wave.

Substitute $2 Hz$ for $f$ in the above expression.

$t=12 Hz=0.500 s$

Conclusion:

Therefore, the period of the wave is $0.500 s$ .

(E)

Expert Solution

To determine

The speed of propagation of wave.

Answer to Problem 16.15P

The speed of propagation of wave is $32.0 m/s$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The formula to calculate speed of propagation of wave is,

$s=fλ$

Here,

$s$ is speed of propagation of wave.

$f$ is frequency of wave.

Substitute $2 s−1$ for $f$ and $16.0 m$ for $λ$ in the above expression.

$s=(2 s−1)(16.0 m)=32.0 m/s$

Conclusion:

Therefore, the speed of propagation of wave is $32.0 m/s$ .

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Answer 2

Textbook Question

Chapter 16, Problem 16.15P

A transverse wave on a siring is described by the wave function

$y = 0.120 sin ( π 8 x + 4 π t )$

where x and y are in meters and t is in seconds. Determine (a) the transverse speed and (b) the transverse acceleration at t = 0.200 s for an element of the string located at x = 1.60 m. What are (c) the wavelength, (d) the period, and (e) the speed of propagation of this wave?

(a)

Expert Solution

To determine

The transverse speed for an element located at $1.60 m$ .

Answer to Problem 16.15P

The transverse speed for an element located at $1.60 m$ is $−1.51 m/s$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The standard equation of the transverse wave is,

$y=Asin(kx−ωt)$ . (1)

Here,

$A$ is amplitude of wave.

$k$ is angular wave number of wave.

$ω$ is angular velocity of wave.

The wave function give is,

$y=0.120sin(π8x+4πt)$ . (2)

Compare the equation (1) and equation (2).

$A=0.120 mk=π8ω=4π rad/s$

The formula to calculate frequency is,

$f=ω2π$

Substitute $4π rad/s$ for $ω$ in the above expression.

$f=4π rad/s2π=2.00 Hz$

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

$v=dydt$

Here,

$dy$ is change in position.

$dt$ is change in time.

Substitute $0.120sin(π8x+4πt)$ for $y$ in the above expression.

$v=d(0.120sin(π8x+4πt))dt$

Differentiate and solve the above expression for $v$ ,

$=(0.120)dsin(π8x+4πt)+sin(π8x+4πt)d(0.120)dtdt=(0.120)cos(π8x+4πt)(4π)+0$ (1)

Substitute $0.200 s$ for $t$ and $1.6 m$ for $x$ in the above expression.

$=(0.120)cos(π8(1.6 m)+4π(0.200 s))(4π)+0=(0.120)(4π)cos(0.2π+0.8π)=−1.5072 m/s≈−1.51 m/s$

Conclusion:

Therefore, the transverse speed for an element located at $1.60 m$ is $−1.51 m/s$ .

(b)

Expert Solution

To determine

To write: The transverse acceleration of the wave.

Answer to Problem 16.15P

The transverse acceleration of the wave is $0$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ . The time of the wave is $0.200 s$ and the position of element is $1.60 m$ .

The change in velocity with respect to time gives the acceleration.

$a=dvdt$ (2)

Here,

$dv$ is change in velocity.

$dt$ is change in time.

From equation (1), the speed is,

$v=(0.120)cos(π8x+4πt)(4π)+0$

Substitute $(0.120)cos(π8x+4πt)(4π)+0$ for $v$ in equation (2).

$a=d(0.120)cos(π8x+4πt)(4π)+0dt$

Differentiate and solve the above expression for $a$ .

$a=d(0.120)cos(π8x+4πt)(4π)+0dt=(0.120)×(−sin((π8x+4πt))×(4π)2)+0$

Substitute $0.200 s$ for $t$ and $1.60 m$ for $x$ in the above expression.

$(0.120)×(−sin((π8(1.60 m)+4π(0.200 s)))×(4π)2)+0$

Solve the above expression.

$=(0.120)×(−sin((π8(1.60 m)+4π(0.200 s)))×(4π)2)+0=−(0.120)(4π)2(−sin(π))=0$

Conclusion:

Therefore, the transverse acceleration of the wave is $0$ .

(c)

Expert Solution

To determine

The wavelength of the wave.

Answer to Problem 16.15P

The wavelength of the wave is $16.0 m$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The formula to calculate wavelength of the wave is,

$λ=2πk$

Here,

$k$ is angular wave number of wave.

Substitute $π8$ for $k$ in the above expression.

$λ=2π(π8)$

Solve the above expression for $λ$ .

$λ=2π(π8)=2×8 m=16.0 m$

Conclusion:

Therefore, the wavelength of the wave is $16.0 m$ .

(d)

Expert Solution

To determine

The period of the wave.

Answer to Problem 16.15P

The period of the wave is $0.500 s$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The formula to calculate frequency is,

$f=ω2π$

Here,

$ω$ is angular frequency of wave.

Substitute $4π rad/s$ for $ω$ in the above expression.

$f=4π rad/s2π=2 Hz$

The formula to calculate time period of the wave is,

$t=1f$

Here,

$f$ is frequency of wave.

Substitute $2 Hz$ for $f$ in the above expression.

$t=12 Hz=0.500 s$

Conclusion:

Therefore, the period of the wave is $0.500 s$ .

(E)

Expert Solution

To determine

The speed of propagation of wave.

Answer to Problem 16.15P

The speed of propagation of wave is $32.0 m/s$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The formula to calculate speed of propagation of wave is,

$s=fλ$

Here,

$s$ is speed of propagation of wave.

$f$ is frequency of wave.

Substitute $2 s−1$ for $f$ and $16.0 m$ for $λ$ in the above expression.

$s=(2 s−1)(16.0 m)=32.0 m/s$

Conclusion:

Therefore, the speed of propagation of wave is $32.0 m/s$ .

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Answer 3

Textbook Question

Chapter 16, Problem 16.15P

A transverse wave on a siring is described by the wave function

$y = 0.120 sin ( π 8 x + 4 π t )$

where x and y are in meters and t is in seconds. Determine (a) the transverse speed and (b) the transverse acceleration at t = 0.200 s for an element of the string located at x = 1.60 m. What are (c) the wavelength, (d) the period, and (e) the speed of propagation of this wave?

(a)

Expert Solution

To determine

The transverse speed for an element located at $1.60 m$ .

Answer to Problem 16.15P

The transverse speed for an element located at $1.60 m$ is $−1.51 m/s$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The standard equation of the transverse wave is,

$y=Asin(kx−ωt)$ . (1)

Here,

$A$ is amplitude of wave.

$k$ is angular wave number of wave.

$ω$ is angular velocity of wave.

The wave function give is,

$y=0.120sin(π8x+4πt)$ . (2)

Compare the equation (1) and equation (2).

$A=0.120 mk=π8ω=4π rad/s$

The formula to calculate frequency is,

$f=ω2π$

Substitute $4π rad/s$ for $ω$ in the above expression.

$f=4π rad/s2π=2.00 Hz$

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

$v=dydt$

Here,

$dy$ is change in position.

$dt$ is change in time.

Substitute $0.120sin(π8x+4πt)$ for $y$ in the above expression.

$v=d(0.120sin(π8x+4πt))dt$

Differentiate and solve the above expression for $v$ ,

$=(0.120)dsin(π8x+4πt)+sin(π8x+4πt)d(0.120)dtdt=(0.120)cos(π8x+4πt)(4π)+0$ (1)

Substitute $0.200 s$ for $t$ and $1.6 m$ for $x$ in the above expression.

$=(0.120)cos(π8(1.6 m)+4π(0.200 s))(4π)+0=(0.120)(4π)cos(0.2π+0.8π)=−1.5072 m/s≈−1.51 m/s$

Conclusion:

Therefore, the transverse speed for an element located at $1.60 m$ is $−1.51 m/s$ .

(b)

Expert Solution

To determine

To write: The transverse acceleration of the wave.

Answer to Problem 16.15P

The transverse acceleration of the wave is $0$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ . The time of the wave is $0.200 s$ and the position of element is $1.60 m$ .

The change in velocity with respect to time gives the acceleration.

$a=dvdt$ (2)

Here,

$dv$ is change in velocity.

$dt$ is change in time.

From equation (1), the speed is,

$v=(0.120)cos(π8x+4πt)(4π)+0$

Substitute $(0.120)cos(π8x+4πt)(4π)+0$ for $v$ in equation (2).

$a=d(0.120)cos(π8x+4πt)(4π)+0dt$

Differentiate and solve the above expression for $a$ .

$a=d(0.120)cos(π8x+4πt)(4π)+0dt=(0.120)×(−sin((π8x+4πt))×(4π)2)+0$

Substitute $0.200 s$ for $t$ and $1.60 m$ for $x$ in the above expression.

$(0.120)×(−sin((π8(1.60 m)+4π(0.200 s)))×(4π)2)+0$

Solve the above expression.

$=(0.120)×(−sin((π8(1.60 m)+4π(0.200 s)))×(4π)2)+0=−(0.120)(4π)2(−sin(π))=0$

Conclusion:

Therefore, the transverse acceleration of the wave is $0$ .

(c)

Expert Solution

To determine

The wavelength of the wave.

Answer to Problem 16.15P

The wavelength of the wave is $16.0 m$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The formula to calculate wavelength of the wave is,

$λ=2πk$

Here,

$k$ is angular wave number of wave.

Substitute $π8$ for $k$ in the above expression.

$λ=2π(π8)$

Solve the above expression for $λ$ .

$λ=2π(π8)=2×8 m=16.0 m$

Conclusion:

Therefore, the wavelength of the wave is $16.0 m$ .

(d)

Expert Solution

To determine

The period of the wave.

Answer to Problem 16.15P

The period of the wave is $0.500 s$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The formula to calculate frequency is,

$f=ω2π$

Here,

$ω$ is angular frequency of wave.

Substitute $4π rad/s$ for $ω$ in the above expression.

$f=4π rad/s2π=2 Hz$

The formula to calculate time period of the wave is,

$t=1f$

Here,

$f$ is frequency of wave.

Substitute $2 Hz$ for $f$ in the above expression.

$t=12 Hz=0.500 s$

Conclusion:

Therefore, the period of the wave is $0.500 s$ .

(E)

Expert Solution

To determine

The speed of propagation of wave.

Answer to Problem 16.15P

The speed of propagation of wave is $32.0 m/s$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The formula to calculate speed of propagation of wave is,

$s=fλ$

Here,

$s$ is speed of propagation of wave.

$f$ is frequency of wave.

Substitute $2 s−1$ for $f$ and $16.0 m$ for $λ$ in the above expression.

$s=(2 s−1)(16.0 m)=32.0 m/s$

Conclusion:

Therefore, the speed of propagation of wave is $32.0 m/s$ .

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Answer 4

Textbook Question

Chapter 16, Problem 16.15P

A transverse wave on a siring is described by the wave function

$y = 0.120 sin ( π 8 x + 4 π t )$

where x and y are in meters and t is in seconds. Determine (a) the transverse speed and (b) the transverse acceleration at t = 0.200 s for an element of the string located at x = 1.60 m. What are (c) the wavelength, (d) the period, and (e) the speed of propagation of this wave?

(a)

Expert Solution

To determine

The transverse speed for an element located at $1.60 m$ .

Answer to Problem 16.15P

The transverse speed for an element located at $1.60 m$ is $−1.51 m/s$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The standard equation of the transverse wave is,

$y=Asin(kx−ωt)$ . (1)

Here,

$A$ is amplitude of wave.

$k$ is angular wave number of wave.

$ω$ is angular velocity of wave.

The wave function give is,

$y=0.120sin(π8x+4πt)$ . (2)

Compare the equation (1) and equation (2).

$A=0.120 mk=π8ω=4π rad/s$

The formula to calculate frequency is,

$f=ω2π$

Substitute $4π rad/s$ for $ω$ in the above expression.

$f=4π rad/s2π=2.00 Hz$

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

$v=dydt$

Here,

$dy$ is change in position.

$dt$ is change in time.

Substitute $0.120sin(π8x+4πt)$ for $y$ in the above expression.

$v=d(0.120sin(π8x+4πt))dt$

Differentiate and solve the above expression for $v$ ,

$=(0.120)dsin(π8x+4πt)+sin(π8x+4πt)d(0.120)dtdt=(0.120)cos(π8x+4πt)(4π)+0$ (1)

Substitute $0.200 s$ for $t$ and $1.6 m$ for $x$ in the above expression.

$=(0.120)cos(π8(1.6 m)+4π(0.200 s))(4π)+0=(0.120)(4π)cos(0.2π+0.8π)=−1.5072 m/s≈−1.51 m/s$

Conclusion:

Therefore, the transverse speed for an element located at $1.60 m$ is $−1.51 m/s$ .

(b)

Expert Solution

To determine

To write: The transverse acceleration of the wave.

Answer to Problem 16.15P

The transverse acceleration of the wave is $0$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ . The time of the wave is $0.200 s$ and the position of element is $1.60 m$ .

The change in velocity with respect to time gives the acceleration.

$a=dvdt$ (2)

Here,

$dv$ is change in velocity.

$dt$ is change in time.

From equation (1), the speed is,

$v=(0.120)cos(π8x+4πt)(4π)+0$

Substitute $(0.120)cos(π8x+4πt)(4π)+0$ for $v$ in equation (2).

$a=d(0.120)cos(π8x+4πt)(4π)+0dt$

Differentiate and solve the above expression for $a$ .

$a=d(0.120)cos(π8x+4πt)(4π)+0dt=(0.120)×(−sin((π8x+4πt))×(4π)2)+0$

Substitute $0.200 s$ for $t$ and $1.60 m$ for $x$ in the above expression.

$(0.120)×(−sin((π8(1.60 m)+4π(0.200 s)))×(4π)2)+0$

Solve the above expression.

$=(0.120)×(−sin((π8(1.60 m)+4π(0.200 s)))×(4π)2)+0=−(0.120)(4π)2(−sin(π))=0$

Conclusion:

Therefore, the transverse acceleration of the wave is $0$ .

(c)

Expert Solution

To determine

The wavelength of the wave.

Answer to Problem 16.15P

The wavelength of the wave is $16.0 m$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The formula to calculate wavelength of the wave is,

$λ=2πk$

Here,

$k$ is angular wave number of wave.

Substitute $π8$ for $k$ in the above expression.

$λ=2π(π8)$

Solve the above expression for $λ$ .

$λ=2π(π8)=2×8 m=16.0 m$

Conclusion:

Therefore, the wavelength of the wave is $16.0 m$ .

(d)

Expert Solution

To determine

The period of the wave.

Answer to Problem 16.15P

The period of the wave is $0.500 s$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The formula to calculate frequency is,

$f=ω2π$

Here,

$ω$ is angular frequency of wave.

Substitute $4π rad/s$ for $ω$ in the above expression.

$f=4π rad/s2π=2 Hz$

The formula to calculate time period of the wave is,

$t=1f$

Here,

$f$ is frequency of wave.

Substitute $2 Hz$ for $f$ in the above expression.

$t=12 Hz=0.500 s$

Conclusion:

Therefore, the period of the wave is $0.500 s$ .

(E)

Expert Solution

To determine

The speed of propagation of wave.

Answer to Problem 16.15P

The speed of propagation of wave is $32.0 m/s$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The formula to calculate speed of propagation of wave is,

$s=fλ$

Here,

$s$ is speed of propagation of wave.

$f$ is frequency of wave.

Substitute $2 s−1$ for $f$ and $16.0 m$ for $λ$ in the above expression.

$s=(2 s−1)(16.0 m)=32.0 m/s$

Conclusion:

Therefore, the speed of propagation of wave is $32.0 m/s$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The transverse speed for an element located at $1.60 m$ .

Answer to Problem 16.15P

The transverse speed for an element located at $1.60 m$ is $−1.51 m/s$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The standard equation of the transverse wave is,

$y=Asin(kx−ωt)$ . (1)

Here,

$A$ is amplitude of wave.

$k$ is angular wave number of wave.

$ω$ is angular velocity of wave.

The wave function give is,

$y=0.120sin(π8x+4πt)$ . (2)

Compare the equation (1) and equation (2).

$A=0.120 mk=π8ω=4π rad/s$

The formula to calculate frequency is,

$f=ω2π$

Substitute $4π rad/s$ for $ω$ in the above expression.

$f=4π rad/s2π=2.00 Hz$

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

$v=dydt$

Here,

$dy$ is change in position.

$dt$ is change in time.

Substitute $0.120sin(π8x+4πt)$ for $y$ in the above expression.

$v=d(0.120sin(π8x+4πt))dt$

Differentiate and solve the above expression for $v$ ,

$=(0.120)dsin(π8x+4πt)+sin(π8x+4πt)d(0.120)dtdt=(0.120)cos(π8x+4πt)(4π)+0$ (1)

Substitute $0.200 s$ for $t$ and $1.6 m$ for $x$ in the above expression.

$=(0.120)cos(π8(1.6 m)+4π(0.200 s))(4π)+0=(0.120)(4π)cos(0.2π+0.8π)=−1.5072 m/s≈−1.51 m/s$

Conclusion:

Therefore, the transverse speed for an element located at $1.60 m$ is $−1.51 m/s$ .

(b)

Expert Solution

To determine

To write: The transverse acceleration of the wave.

Answer to Problem 16.15P

The transverse acceleration of the wave is $0$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ . The time of the wave is $0.200 s$ and the position of element is $1.60 m$ .

The change in velocity with respect to time gives the acceleration.

$a=dvdt$ (2)

Here,

$dv$ is change in velocity.

$dt$ is change in time.

From equation (1), the speed is,

$v=(0.120)cos(π8x+4πt)(4π)+0$

Substitute $(0.120)cos(π8x+4πt)(4π)+0$ for $v$ in equation (2).

$a=d(0.120)cos(π8x+4πt)(4π)+0dt$

Differentiate and solve the above expression for $a$ .

$a=d(0.120)cos(π8x+4πt)(4π)+0dt=(0.120)×(−sin((π8x+4πt))×(4π)2)+0$

Substitute $0.200 s$ for $t$ and $1.60 m$ for $x$ in the above expression.

$(0.120)×(−sin((π8(1.60 m)+4π(0.200 s)))×(4π)2)+0$

Solve the above expression.

$=(0.120)×(−sin((π8(1.60 m)+4π(0.200 s)))×(4π)2)+0=−(0.120)(4π)2(−sin(π))=0$

Conclusion:

Therefore, the transverse acceleration of the wave is $0$ .

(c)

Expert Solution

To determine

The wavelength of the wave.

Answer to Problem 16.15P

The wavelength of the wave is $16.0 m$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The formula to calculate wavelength of the wave is,

$λ=2πk$

Here,

$k$ is angular wave number of wave.

Substitute $π8$ for $k$ in the above expression.

$λ=2π(π8)$

Solve the above expression for $λ$ .

$λ=2π(π8)=2×8 m=16.0 m$

Conclusion:

Therefore, the wavelength of the wave is $16.0 m$ .

(d)

Expert Solution

To determine

The period of the wave.

Answer to Problem 16.15P

The period of the wave is $0.500 s$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The formula to calculate frequency is,

$f=ω2π$

Here,

$ω$ is angular frequency of wave.

Substitute $4π rad/s$ for $ω$ in the above expression.

$f=4π rad/s2π=2 Hz$

The formula to calculate time period of the wave is,

$t=1f$

Here,

$f$ is frequency of wave.

Substitute $2 Hz$ for $f$ in the above expression.

$t=12 Hz=0.500 s$

Conclusion:

Therefore, the period of the wave is $0.500 s$ .

(E)

Expert Solution

To determine

The speed of propagation of wave.

Answer to Problem 16.15P

The speed of propagation of wave is $32.0 m/s$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The formula to calculate speed of propagation of wave is,

$s=fλ$

Here,

$s$ is speed of propagation of wave.

$f$ is frequency of wave.

Substitute $2 s−1$ for $f$ and $16.0 m$ for $λ$ in the above expression.

$s=(2 s−1)(16.0 m)=32.0 m/s$

Conclusion:

Therefore, the speed of propagation of wave is $32.0 m/s$ .

Answer 8

(a)

Expert Solution

To determine

The transverse speed for an element located at $1.60 m$ .

Answer to Problem 16.15P

The transverse speed for an element located at $1.60 m$ is $−1.51 m/s$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The standard equation of the transverse wave is,

$y=Asin(kx−ωt)$ . (1)

Here,

$A$ is amplitude of wave.

$k$ is angular wave number of wave.

$ω$ is angular velocity of wave.

The wave function give is,

$y=0.120sin(π8x+4πt)$ . (2)

Compare the equation (1) and equation (2).

$A=0.120 mk=π8ω=4π rad/s$

The formula to calculate frequency is,

$f=ω2π$

Substitute $4π rad/s$ for $ω$ in the above expression.

$f=4π rad/s2π=2.00 Hz$

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

$v=dydt$

Here,

$dy$ is change in position.

$dt$ is change in time.

Substitute $0.120sin(π8x+4πt)$ for $y$ in the above expression.

$v=d(0.120sin(π8x+4πt))dt$

Differentiate and solve the above expression for $v$ ,

$=(0.120)dsin(π8x+4πt)+sin(π8x+4πt)d(0.120)dtdt=(0.120)cos(π8x+4πt)(4π)+0$ (1)

Substitute $0.200 s$ for $t$ and $1.6 m$ for $x$ in the above expression.

$=(0.120)cos(π8(1.6 m)+4π(0.200 s))(4π)+0=(0.120)(4π)cos(0.2π+0.8π)=−1.5072 m/s≈−1.51 m/s$

Conclusion:

Therefore, the transverse speed for an element located at $1.60 m$ is $−1.51 m/s$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The transverse speed for an element located at $1.60 m$ .

Answer 13

Answer to Problem 16.15P

The transverse speed for an element located at $1.60 m$ is $−1.51 m/s$ .

Answer 14

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The standard equation of the transverse wave is,

$y=Asin(kx−ωt)$ . (1)

Here,

$A$ is amplitude of wave.

$k$ is angular wave number of wave.

$ω$ is angular velocity of wave.

The wave function give is,

$y=0.120sin(π8x+4πt)$ . (2)

Compare the equation (1) and equation (2).

$A=0.120 mk=π8ω=4π rad/s$

The formula to calculate frequency is,

$f=ω2π$

Substitute $4π rad/s$ for $ω$ in the above expression.

$f=4π rad/s2π=2.00 Hz$

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

$v=dydt$

Here,

$dy$ is change in position.

$dt$ is change in time.

Substitute $0.120sin(π8x+4πt)$ for $y$ in the above expression.

$v=d(0.120sin(π8x+4πt))dt$

Differentiate and solve the above expression for $v$ ,

$=(0.120)dsin(π8x+4πt)+sin(π8x+4πt)d(0.120)dtdt=(0.120)cos(π8x+4πt)(4π)+0$ (1)

Substitute $0.200 s$ for $t$ and $1.6 m$ for $x$ in the above expression.

$=(0.120)cos(π8(1.6 m)+4π(0.200 s))(4π)+0=(0.120)(4π)cos(0.2π+0.8π)=−1.5072 m/s≈−1.51 m/s$

Conclusion:

Therefore, the transverse speed for an element located at $1.60 m$ is $−1.51 m/s$ .

Answer 15

(b)

Expert Solution

To determine

To write: The transverse acceleration of the wave.

Answer to Problem 16.15P

The transverse acceleration of the wave is $0$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ . The time of the wave is $0.200 s$ and the position of element is $1.60 m$ .

The change in velocity with respect to time gives the acceleration.

$a=dvdt$ (2)

Here,

$dv$ is change in velocity.

$dt$ is change in time.

From equation (1), the speed is,

$v=(0.120)cos(π8x+4πt)(4π)+0$

Substitute $(0.120)cos(π8x+4πt)(4π)+0$ for $v$ in equation (2).

$a=d(0.120)cos(π8x+4πt)(4π)+0dt$

Differentiate and solve the above expression for $a$ .

$a=d(0.120)cos(π8x+4πt)(4π)+0dt=(0.120)×(−sin((π8x+4πt))×(4π)2)+0$

Substitute $0.200 s$ for $t$ and $1.60 m$ for $x$ in the above expression.

$(0.120)×(−sin((π8(1.60 m)+4π(0.200 s)))×(4π)2)+0$

Solve the above expression.

$=(0.120)×(−sin((π8(1.60 m)+4π(0.200 s)))×(4π)2)+0=−(0.120)(4π)2(−sin(π))=0$

Conclusion:

Therefore, the transverse acceleration of the wave is $0$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

To write: The transverse acceleration of the wave.

Answer 20

Answer to Problem 16.15P

The transverse acceleration of the wave is $0$ .

Answer 21

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ . The time of the wave is $0.200 s$ and the position of element is $1.60 m$ .

The change in velocity with respect to time gives the acceleration.

$a=dvdt$ (2)

Here,

$dv$ is change in velocity.

$dt$ is change in time.

From equation (1), the speed is,

$v=(0.120)cos(π8x+4πt)(4π)+0$

Substitute $(0.120)cos(π8x+4πt)(4π)+0$ for $v$ in equation (2).

$a=d(0.120)cos(π8x+4πt)(4π)+0dt$

Differentiate and solve the above expression for $a$ .

$a=d(0.120)cos(π8x+4πt)(4π)+0dt=(0.120)×(−sin((π8x+4πt))×(4π)2)+0$

Substitute $0.200 s$ for $t$ and $1.60 m$ for $x$ in the above expression.

$(0.120)×(−sin((π8(1.60 m)+4π(0.200 s)))×(4π)2)+0$

Solve the above expression.

$=(0.120)×(−sin((π8(1.60 m)+4π(0.200 s)))×(4π)2)+0=−(0.120)(4π)2(−sin(π))=0$

Conclusion:

Therefore, the transverse acceleration of the wave is $0$ .

Answer 22

(c)

Expert Solution

To determine

The wavelength of the wave.

Answer to Problem 16.15P

The wavelength of the wave is $16.0 m$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The formula to calculate wavelength of the wave is,

$λ=2πk$

Here,

$k$ is angular wave number of wave.

Substitute $π8$ for $k$ in the above expression.

$λ=2π(π8)$

Solve the above expression for $λ$ .

$λ=2π(π8)=2×8 m=16.0 m$

Conclusion:

Therefore, the wavelength of the wave is $16.0 m$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The wavelength of the wave.

Answer 27

Answer to Problem 16.15P

The wavelength of the wave is $16.0 m$ .

Answer 28

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The formula to calculate wavelength of the wave is,

$λ=2πk$

Here,

$k$ is angular wave number of wave.

Substitute $π8$ for $k$ in the above expression.

$λ=2π(π8)$

Solve the above expression for $λ$ .

$λ=2π(π8)=2×8 m=16.0 m$

Conclusion:

Therefore, the wavelength of the wave is $16.0 m$ .

Answer 29

(d)

Expert Solution

To determine

The period of the wave.

Answer to Problem 16.15P

The period of the wave is $0.500 s$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The formula to calculate frequency is,

$f=ω2π$

Here,

$ω$ is angular frequency of wave.

Substitute $4π rad/s$ for $ω$ in the above expression.

$f=4π rad/s2π=2 Hz$

The formula to calculate time period of the wave is,

$t=1f$

Here,

$f$ is frequency of wave.

Substitute $2 Hz$ for $f$ in the above expression.

$t=12 Hz=0.500 s$

Conclusion:

Therefore, the period of the wave is $0.500 s$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

The period of the wave.

Answer 34

Answer to Problem 16.15P

The period of the wave is $0.500 s$ .

Answer 35

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The formula to calculate frequency is,

$f=ω2π$

Here,

$ω$ is angular frequency of wave.

Substitute $4π rad/s$ for $ω$ in the above expression.

$f=4π rad/s2π=2 Hz$

The formula to calculate time period of the wave is,

$t=1f$

Here,

$f$ is frequency of wave.

Substitute $2 Hz$ for $f$ in the above expression.

$t=12 Hz=0.500 s$

Conclusion:

Therefore, the period of the wave is $0.500 s$ .

Answer 36

(E)

Expert Solution

To determine

The speed of propagation of wave.

Answer to Problem 16.15P

The speed of propagation of wave is $32.0 m/s$ .

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The formula to calculate speed of propagation of wave is,

$s=fλ$

Here,

$s$ is speed of propagation of wave.

$f$ is frequency of wave.

Substitute $2 s−1$ for $f$ and $16.0 m$ for $λ$ in the above expression.

$s=(2 s−1)(16.0 m)=32.0 m/s$

Conclusion:

Therefore, the speed of propagation of wave is $32.0 m/s$ .

Answer 37

(E)

Expert Solution

Answer 38

(E)

Expert Solution

Answer 39

Expert Solution

Answer 40

To determine

The speed of propagation of wave.

Answer 41

Answer to Problem 16.15P

The speed of propagation of wave is $32.0 m/s$ .

Answer 42

Explanation of Solution

Given info: The wave function of the wave is $y=0.120sin(π8x+4πt)$ .

The formula to calculate speed of propagation of wave is,

$s=fλ$

Here,

$s$ is speed of propagation of wave.

$f$ is frequency of wave.

Substitute $2 s−1$ for $f$ and $16.0 m$ for $λ$ in the above expression.

$s=(2 s−1)(16.0 m)=32.0 m/s$

Conclusion:

Therefore, the speed of propagation of wave is $32.0 m/s$ .

Answer 43

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