First slide

Simple hormonic motion

Question

A 4 kg particle is moving along the x-axis under the action of the force $F = - (\frac{π^{2}}{16}) xN$ . At t = 2 sec, the particle passes through the origin and at t = l0 sec its speed is $4 \sqrt{2}$ m/s. The amplitude of the motion is:

Moderate

A

$\frac{32 \sqrt{2}}{π} m$
B

$\frac{16}{π} m$
C

$\frac{4}{π} m$
D

$\frac{16 \sqrt{2}}{π} m$

Solution

$a = - (\frac{π^{2}}{64}) x \Rightarrow ω = \sqrt{\frac{π^{2}}{64}} = \frac{π}{8}$
$\Rightarrow T = \frac{2 π}{ω} = 16 \sec .$
There is a time difference of $\frac{T}{2}$ between t = 2 sec to t = l0 sec. Hence particle is again passing through the mean position of SHM where its speed is maximum.
i.e., $V_{\max} = Aω = 4 \sqrt{2}$
$\Rightarrow A = \frac{4 \sqrt{2}}{\frac{π}{8}} = \frac{32 \sqrt{2}}{π} m .$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App