Question

A wave on a string is described by $\mathit{y}\left(x,t\right)\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\left(\pi x/8-4\pi t\right)$, where xand yare in centimeters and tis in seconds. (a) What is the transverse speed for a point on the string at x = 6.00 cm when t = 0.250 s? (b) What is the maximum transverse speed of any point on the string? (c) What is the magnitude of the transverse acceleration for a point on the string at x = 6.00 cm when t = 0.250 s? (d) What is the magnitude of the maximum transverse acceleration for any point on the string?

Short answer

a) The magnitude of the transverse speed for a point on the string at x = 6.00cm when $t=0.250s$is 1.33 m/s

b) The maximum transverse speed of any point on the string is 1.88m/s

c) The magnitude of the transverse acceleration for a point on the string at x=6.00cm when t =0.250 is$16.7\mathrm{m}/{\mathrm{s}}^2$

d) The magnitude of the maximum transverse acceleration for any point on the string is$23.7\mathrm{m}/{\mathrm{s}}^2$

Step-by-step solution

Identification of given data

Wave equation,$y\left(x,t\right)=15.0\sin \left(\frac{\mathrm{\pi x}}{8}-4\mathrm{\pi t}\right)$

Understanding the concept of the equation of the wave

Using the equation of velocity of the wave, we can find the magnitude of the transverse speed for a point on the string at when the maximum transverse speed of any point on the string. By differentiating the equation of velocity for t, we get the equation for the acceleration of the wave. From this, we can find the magnitude of the transverse acceleration for a point on the string when and the magnitude of the maximum transverse acceleration for any point on the string.

Formula:

The general expression of the wave,$y\left(x,t\right)=y_m\sin \left(kx-\omega t\right)$ …(i)

The transverse velocity of the wave,$u=dy/dt$ …(ii)

The transverse acceleration of the wave, $a=du/dt$ …(iii)

(a) Determining the magnitude of the transverse speed

We are given that

$y\left(x,t\right)=15.0\sin \left(\frac{\mathrm{\pi x}}{8}-4\mathrm{\pi t}\right)$

Hence, the transverse velocity using equation (ii) is given as:

$$\begin{gathered} u=\frac{d}{dt}\left[15.0\sin \left(\frac{\mathrm{\pi x}}{8}-4\mathrm{\pi t}\right)\right] \\ =15.0\left(4\mathrm{\pi }\right)\cos \left(\frac{\mathrm{\pi x}}{8}\mathrm{\pi t}\right) \end{gathered}$$

At x=6.00cm and$\mathrm{t}=0.250\mathrm{s}\mathrm{or}\frac{1}{4}\mathrm{s}$
$$\begin{gathered} u=-60\mathrm{\pi cos}\left[\frac{\mathrm{\pi }\left(6.00\mathrm{cm}\right)}{8}-4\mathrm{\pi }\left(\frac{1}{4}\mathrm{s}\right)\right] \\ =-60\mathrm{\pi cos}\left(-\frac{\mathrm{\pi }}{4}\right) \\ =-133\frac{\mathrm{cm}}{\mathrm{s}} \\ =1.33\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Therefore, the magnitude of the transverse speed for a point on the string at x=6.00cm when $t=0.250s$is 1.33m/s

(b) Determining the maximum transverse speed

We know for maximum speed,

$$\begin{gathered} u_m=\omega y_m \\ =60\mathrm{\pi } \\ =188.4\frac{\mathrm{cm}}{\mathrm{s}} \\ =1.88\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Therefore, the maximum transverse speed of any point on the string is$1.88\frac{m}{s}$

(c) Determining the magnitude of transverse acceleration

Using equation (iii), we get the magnitude of transverse acceleration as:

$$\begin{gathered} a=\frac{d}{dt}\left[-60\mathrm{\pi cos}\left(\frac{\mathrm{\pi x}}{8}-4\mathrm{\pi t}\right)\right] \\ =-60\mathrm{\pi }\left(4\mathrm{\pi }\right)\sin \left(\frac{\mathrm{\pi x}}{8}-4\mathrm{\pi t}\right) \end{gathered}$$

Hence, at x=6.00cm and role="math" localid="1660997687638" $\mathrm{t}=0.250\mathrm{s}\mathrm{or}\frac{1}{4}\mathrm{s}$,

$$\begin{gathered} a=-240{\mathrm{\pi }}^2\sin \left[\frac{\mathrm{\pi }\left(6.00\mathrm{cm}\right)}{8}-4\mathrm{\pi }\left(\frac{1}{4}\mathrm{s}\right)\right] \\ =-240{\mathrm{\pi }}^2\sin \left(-\frac{\mathrm{\pi }}{4}\right) \\ =16.7\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the magnitude of the transverse acceleration for a point on the string at x=6.00cm when t=0.250s is$16.7\mathrm{m}/{\mathrm{s}}^2$

(d) Determining the maximum transverse acceleration

We know for maximum acceleration,

$$\begin{gathered} a_{\omega }=-{\omega }^2{u}_m \\ =-240{\mathrm{\pi }}^2 \\ =-2368.7\frac{\mathrm{cm}}{{\mathrm{s}}^2} \\ =-23.7\frac{\mathrm{m}}{{\mathrm{s}}^2} \end{gathered}$$

Therefore, magnitude of the maximum transverse acceleration for any point on the string is$-23.7\frac{m}{s^2}$