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?

• 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 .

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)

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 .

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 $1.0\mathrm{m}$.

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 .

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 $2.0\mathrm{m}/\mathrm{s}$.

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 $kx-\omega t$.

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 $0.75\mathrm{m}/\mathrm{s}$.

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 $-0.02\mathrm{m}$.