Chapter 11, Problem 83P

(a)

To determine

Identify the direction of propagation of wave.

Answer to Problem 83P

Wave propagates in left direction.

Explanation of Solution

The equation of wave is $y\left(x,t\right)=\left(7.00\text{cm}\right)\cos \left(\left(6.00\pi \text{rad}/\text{cm}\right)x+\left(20.0\text{rad}/\text{s}\right)t\right)$.

The direction of wave is determined by whether the argument of cosine function is positive or not. If it is positive, means that the wave is propagating in left direction and negative value indicates the propagation in right direction. For the wave $y\left(x,t\right)=\left(7.00\text{cm}\right)\cos \left(\left(6.00\pi \text{rad}/\text{cm}\right)x+\left(20.0\text{rad}/\text{s}\right)t\right)$, argument is positive.

Therefore, the wave propagates in left direction.

(b)

To determine

Identify the maximum distance up to which the particles in the medium will move from the equilibrium position.

Answer to Problem 83P

Particles can move by $7.00\text{cm}$ from the equilibrium position.

Explanation of Solution

The equation of wave is $y\left(x,t\right)=\left(7.00\text{cm}\right)\cos \left(\left(6.00\pi \text{rad}/\text{cm}\right)x+\left(20.0\text{rad}/\text{s}\right)t\right)$.

Write the standard form of equation for a wave moving along –ve x-axis.

$y\left(x,t\right)=A\cos \left(kx+\omega t\right)$

Here, $y\left(x,t\right)$ denotes the displacement of particle at any instant $t$, $A$ is the amplitude, $\omega$ is the angular frequency, and $k$ is the wave number.

The maximum distance up to which the particles in the medium will move from the equilibrium position is called the amplitude of a wave.

Conclusion:

Compare $y\left(x,t\right)=\left(7.00\text{cm}\right)\cos \left(\left(6.00\pi \text{rad}/\text{cm}\right)x+\left(20.0\pi \text{rad}/\text{s}\right)t\right)$ with the standard form.

Write the value of $A$.

$A=7.00\text{cm}$

Therefore, the particles can move by $7.00\text{cm}$ from the equilibrium position.

(c)

To determine

Frequency of wave.

Answer to Problem 83P

Frequency is $10.0\text{Hz}$.

Explanation of Solution

The equation of wave is $y\left(x,t\right)=\left(7.00\text{cm}\right)\cos \left(\left(6.00\pi \text{rad}/\text{cm}\right)x+\left(20.0\pi \text{rad}/\text{s}\right)t\right)$.

Compare $y\left(x,t\right)=\left(7.00\text{cm}\right)\cos \left(\left(6.00\pi \text{rad}/\text{cm}\right)x+\left(20.0\text{rad}/\text{s}\right)t\right)$ with the standard form $y\left(x,t\right)=A\cos \left(kx+\omega t\right)$.

Write the value of $\omega$.

$\omega =20.0\pi \text{rad}/\text{s}$

Write the relation between the linear frequency and $\omega$.

$f=\frac{\omega }{2\pi }$

Here, $f$ is the frequency of wave.

Conclusion:

Substitute $20.0\pi \text{rad}/\text{s}$ for $\omega$ in the above equation to find $f$.

$\begin{array}{c}f=\frac{20.0\pi \text{rad}/\text{s}}{2\pi }\\ =10\pi \text{Hz}\end{array}$

Therefore, the frequency is $10.0\text{Hz}$.

(d)

To determine

Wavelength of wave.

Answer to Problem 83P

Wavelength is $0.333\text{cm}$.

Explanation of Solution

The equation of wave is $y\left(x,t\right)=\left(7.00\text{cm}\right)\cos \left(\left(6.00\pi \text{rad}/\text{cm}\right)x+\left(20.0\pi \text{rad}/\text{s}\right)t\right)$.

Compare $y\left(x,t\right)=\left(7.00\text{cm}\right)\cos \left(\left(6.00\pi \text{rad}/\text{cm}\right)x+\left(20.0\text{rad}/\text{s}\right)t\right)$ with the standard form $y\left(x,t\right)=A\cos \left(kx+\omega t\right)$.

Write the value of $k$.

$k=6.00\pi \text{rad}/\text{cm}$

Write the relation between wavelength and $k$.

$\lambda =\frac{2\pi }{k}$

Here, $\lambda$ is the wavelength of wave.

Conclusion:

Substitute $6.00\pi \text{rad}/\text{cm}$ for $k$ in the above equation to find $\lambda$.

$\begin{array}{c}\lambda =\frac{2\pi }{6.00\pi \text{rad}/\text{cm}}\\ =0.333\text{cm}\end{array}$

Therefore, the wavelength is $0.333\text{cm}$.

(e)

To determine

Speed of wave.

Answer to Problem 83P

Speed is $3.33\text{cm}/\text{s}$.

Explanation of Solution

The equation of wave is $y\left(x,t\right)=\left(7.00\text{cm}\right)\cos \left(\left(6.00\pi \text{rad}/\text{cm}\right)x+\left(20.0\pi \text{rad}/\text{s}\right)t\right)$.

Write the relation between the speed of wave and $k$.

$v=\frac{\omega }{k}$

Here, $v$ is the speed of wave.

Conclusion:

Substitute $20.0\pi \text{rad}/\text{s}$ for $\omega$ and $6.00\pi \text{rad}/\text{cm}$ for $k$ in the above equation to find $v$.

$\begin{array}{c}v=\frac{20.0\pi \text{rad}/\text{s}}{6.00\pi \text{rad}/\text{cm}}\\ =3.33\text{cm}/\text{s}\end{array}$

Therefore, the speed is $3.33\text{cm}/\text{s}$.

(f)

To determine

Explain the motion at $y=7.00\text{cm}$ and $x=0$ when $t=0$.

Explanation of Solution

The equation of wave is $y\left(x,t\right)=\left(7.00\text{cm}\right)\cos \left(\left(6.00\pi \text{rad}/\text{cm}\right)x+\left(20.0\pi \text{rad}/\text{s}\right)t\right)$.

In the given equation of wave, $y\left(x,t\right)$ denotes the displacement of particle and the variable along with $k$ inside the cosine function denotes the direction of propagation of wave. Cosine function shows that the particle is executing sinusoidal motion. Displacement of particle $y\left(x,t\right)$ means the particle is moving along the y-axis.

On substituting $x=0$ and $t=0$ in the given equation, will results in the maximum value of cosine function means that the particle will be at the maximum amplitude at the moment. Since the given form equation is consistent with stand form of equation, it can be say the particle is vibrating about $y=0$.

Therefore, the particle is under sinusoidal oscillation with amplitude $7.00\text{cm}$ about $y=0$ along y-axis.

(g)

To determine

Check whether the wave is transverse of longitudinal in nature.

Explanation of Solution

Longitudinal wave is one which the displacement of particles of medium and the direction of wave are in same direction. For transverse wave, displacement of particles of medium and the direction of wave will be perpendicular to each other.

In the give wave, it is found that particle is vibrating along y-axis and the wave propagates along x-axis.

Therefore, the given wave is a transverse wave.