Chapter 11, Problem 10P

To determine

To Find: The displacement from the equilibrium.

Answer to Problem 10P

The speed of SHO is half the maximum value of velocity is at $0.866A$ .

Explanation of Solution

Given info:

  $\begin{array}{l}x=A\cos (\omega t)\\ v=\frac{1}{2}\times v_{\max }\end{array}$

Formula used:

Expression of displacement:

  $\begin{array}{l}v=\frac{dx}{dt}\\ v=\frac{1}{2}\times v_{\max }\end{array}$

Here,

  $v$ is the velocity,

  $v_{\text{max}}$ is the maximum velocity,

  $\frac{dx}{dt}$ is the rate of change of displacement.

Calculation:

Substitute the given values in the above equation:

  $x=A\cos (\omega t)$

  $v=\frac{dx}{dt}=-A\times \omega \times \sin \omega t$

Velocity is maximum when $\sin \omega t=1$ ,

  $v=v_{\max }=-A\times \omega$

  $\begin{array}{l}v=\frac{1}{2}\times v_{\max }\\ v=\frac{1}{2}\times (-A\times \omega )\\ -A\times \omega \times \sin \omega t=-\frac{1}{2}\times A\times \omega \\ \sin \omega t=\frac{1}{2}=\sin \frac{\pi }{6}\\ \omega t=\frac{\pi }{6}\end{array}$

Put this value in the following equation to get the displacement,

  $\begin{array}{l}x=A\cos (\omega t)\\ \omega t=\frac{\pi }{6}\\ x=A\cos \frac{\pi }{6}=0.866A\end{array}$

Conclusion:

Thus, the displacement at which the speed of SHO is half the maximum value of velocity is $x=0.866A$ .