91Ó°ÊÓ

Wave equation,$y\left(x,t\right)=15.0\sin \left(\frac{\mathrm{Ï€³\ae }}{8}-4\mathrm{Ï€³Ù}\right)$

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,$y\left(x,t\right)=y_m\sin \left(kx-\mathrm{Ó\neg }t\right)$ …(¾±)

The transverse velocity of the wave,$u=dy/dt$ …(¾±¾±)

The transverse acceleration of the wave, $a=du/dt$ …(¾±¾±¾±)

We are given that

$y\left(x,t\right)=15.0\sin \left(\frac{\mathrm{Ï€³\ae }}{8}-4\mathrm{Ï€³Ù}\right)$

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

$$\begin{gathered} u=\frac{d}{dt}\left[15.0\sin \left(\frac{\mathrm{Ï€³\ae }}{8}-4\mathrm{Ï€³Ù}\right)\right] \\ =15.0\left(4\mathrm{Ï€}\right)\cos \left(\frac{\mathrm{Ï€³\ae }}{8}\mathrm{Ï€³Ù}\right) \end{gathered}$$

At x=6.00cm and$\mathrm{t}=0.250\mathrm{s}\mathrm{or}\frac{1}{4}\mathrm{s}$
$$\begin{gathered} u=-60\mathrm{Ï€³¦´Ç²õ}\left[\frac{\mathrm{Ï€}\left(6.00\mathrm{cm}\right)}{8}-4\mathrm{Ï€}\left(\frac{1}{4}\mathrm{s}\right)\right] \\ =-60\mathrm{Ï€³¦´Ç²õ}\left(-\frac{\mathrm{Ï€}}{4}\right) \\ =-133\frac{\mathrm{cm}}{\mathrm{s}} \\ =1.33\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

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

We know for maximum speed,

$$\begin{gathered} u_m=\mathrm{Ó\neg }{y}_m \\ =60\mathrm{Ï€} \\ =188.4\frac{\mathrm{cm}}{\mathrm{s}} \\ =1.88\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

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

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

$$\begin{gathered} a=\frac{d}{dt}\left[-60\mathrm{Ï€³¦´Ç²õ}\left(\frac{\mathrm{Ï€³\ae }}{8}-4\mathrm{Ï€³Ù}\right)\right] \\ =-60\mathrm{Ï€}\left(4\mathrm{Ï€}\right)\sin \left(\frac{\mathrm{Ï€³\ae }}{8}-4\mathrm{Ï€³Ù}\right) \end{gathered}$$

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

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

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

We know for maximum acceleration,

$$\begin{gathered} a_{\mathrm{Ó\neg }}=-{\mathrm{Ó\neg }}^2{u}_m \\ =-240{\mathrm{Ï€}}^2 \\ =-2368.7\frac{\mathrm{cm}}{{\mathrm{s}}^2} \\ =-23.7\frac{\mathrm{m}}{{\mathrm{s}}^2} \end{gathered}$$

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