First slide

Simple harmonic motion

Question

A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then its periodic time (in seconds) is

Moderate

A

$\frac{4π}{3}$
B

$\frac{8π}{3}$
C

$\frac{7π}{3}$
D

$\frac{3π}{8}$

Solution

At this time (when x=4 cm) velocity and acceleration have same magnitude
$G i v e n A = 5 c m a n d a t d i s p l a c e m e n t x = 4 c m, v e l o c i t y = a c c e l e r a t i o n \Rightarrow |ω \sqrt{A^{2} {-x}^{2}}| = |- ω^{2} x| ⇒ \sqrt{5^{2} {-4}^{2}} =ω(4) \Rightarrow ω = \frac{3}{4} = \frac{2 π}{T} \Rightarrow T = \frac{8 π}{3}$

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