Question

The equation of a transverse wave traveling along a very long string is $\mathit{y}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathit{s}\mathit{i}\mathit{n}\left(0.020\mathrm{\pi x}+4.0\mathrm{\pi t}\right)$, where x andy are expressed in centimeters and is in seconds. (a) Determine the amplitude,(b) Determine the wavelength, (c)Determine the frequency, (d) Determine the speed, (e) Determine the direction of propagation of the wave, and (f) Determine the maximum transverse speed of a particle in the string. (g)What is the transverse displacement atx = 3.5 cmwhen t = 0.26 s?

Short answer

1. The amplitude is, 0.06 m .
2. The wavelength is, 1.0 m .
3. The frequency is, 2.0 Hz .
4. The speed is 2.0 m/s .
5. The direction of propagation of waves is in the $-x$direction.
6. The maximum transverse speed of a particle in the string is, 0.75 m/s .
7. The transverse displacement at x=3.5 cm when t=0.26 s is, $-0.02\mathrm{cm}$.

Step-by-step solution

The given data

• The transverse wave travelling along a very long string, $y=6.0\sin \left(0.020\mathrm{\pi x}+4.0\mathrm{\pi t}\right)$
• Position and time of the wave are given, x = 3.5 cm and t = 0.26 s .

Understanding the concept of wave equation

By using corresponding formulas, we can find the values of the amplitude, wavelength, frequency, and speed, the direction of propagation of the wave, the maximum transverse speed, and the transverse displacement.

Formula:

The wavelength of a wave, $\lambda =\frac{2\mathrm{\pi }}{k}$ (i)

The frequency of a wave, $f=\frac{\omega }{2\pi }$ (ii)

The speed of a wave, $v=\lambda f$ (iii)

The maximum transverse speed of a wave, $u_{max}=2{\mathrm{\pi fy}}_{\mathrm{m}}$ (iv)

The sinusoidal wave moving in positive x direction, $y\left(x,t\right)=y_m\sin \left(kx-\omega t\right)$ (v)

a) Calculations of amplitude

We know that the sinusoidal wave moving in positive x direction is given by:

$y\left(x,t\right)=y_m\sin \left(kx-\omega t\right)$

Hence, the given transverse wave travelling along a very long string is given as:

$y=6.0\sin \left(0.020\mathrm{\pi x}+4.0\mathrm{\pi t}\right)$

Hence, the amplitude of the wave from the equation is,

$y_m=6.0\mathrm{cm}\mathrm{or}0.06\mathrm{m}$

Hence the value of amplitude is 0.06 m .

b) Calculation of wavelength

We know that,

$$\begin{gathered} \frac{2\mathrm{\pi }}{\lambda }=k \\ =0.020\mathrm{\pi }\left(\because \mathrm{from}\mathrm{the}\mathrm{wave}\mathrm{equation}\right) \end{gathered}$$

Therefore, the wavelength using equation (i) and the given values$\lambda$is given as:

$$\begin{gathered} \lambda =\frac{2\mathrm{\pi }}{0.020\mathrm{\pi }} \\ =\frac{2}{0.020} \\ =1.0\times {10}^2\mathrm{cm} \\ =1.0\mathrm{m} \end{gathered}$$

Hence, the value of wavelength is $1.0\mathrm{m}$.

c) Calculation of frequency

We know that,

$$\begin{gathered} 2\mathrm{\pi f}=\mathrm{\omega } \\ =4.0\mathrm{\pi }\left(\mathrm{from}\mathrm{wave}\mathrm{equation}\right) \end{gathered}$$

Hence, the frequency using equation (ii), we get

$$\begin{gathered} f=\frac{4.0\mathrm{\pi }}{2\mathrm{\pi }} \\ =\frac{4.0}{2} \\ =2.0\mathrm{Hz} \end{gathered}$$

Hence, the value of frequency is 2.0 Hz .

d) Calculation of wave speed

Using equation (iii) and the given values, the speed of the wave is given as:

$$\begin{gathered} v=1.0\times {10}^2\times 2.0 \\ =2.0\times {10}^2\mathrm{cm}/\mathrm{s} \\ =2.0\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the wave speed is $2.0\mathrm{m}/\mathrm{s}$.

e) Finding the direction of propagation

We know that the sinusoidal wave moving in positive x direction is given by,

$y\left(x,t\right)=y_m\sin \left(kx-\omega t\right)$

The wave is propagating in the $-x$direction since the angle of trigonometric function is role="math" localid="1661161274816" $kx+\omega t$instead of $kx-\omega t$.

f) Calculation of maximum transverse speed

The maximum transverse speed using equation (iv) and the given values is given by:

$$\begin{gathered} u_{max}=2\times \mathrm{\pi }\times 2.0\times 6.0 \\ =75\mathrm{cm}/\mathrm{s} \\ =0.75\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of maximum transverse speed is $0.75\mathrm{m}/\mathrm{s}$.

g) Calculation of transverse displacement

We know that, the sinusoidal wave moving in positive x direction is given by

$y\left(x,t\right)=y_m\sin \left(kx-\omega t\right)$

From given, we have

$$\begin{gathered} y\left(3.5\mathrm{cm},0.26\mathrm{s}\right)=6.0\sin \left[0.020\mathrm{\pi }\left(3.5\right)+4.0\mathrm{\pi }\left(0.26\right)\right] \\ =6.0\sin \left[3.49°\right] \\ =-2.0\mathrm{cm} \\ =-0.02\mathrm{m} \end{gathered}$$

Hence, the value transverse displacement is $-0.02\mathrm{m}$.