First slide

Simple harmonic motion

Question

A body executing SHM has a maximum velocity 1 m/sec, maximum acceleration of $4 m / s e c^{2}$ . It’s amplitude in meters is

Moderate

A

1
B

0.75
C

0.5
D

0.25

Solution

$G i v e n v e l o c i t y m a x i m u m i s v_{\max} = 1 m/s=ωA, here w=angular velocity,A=amplitude a c c e l e r a t i o n m a x i m u m a_{max} {=4m/s}^{2} {=ω}^{2} A d i v i d e a b o v e t w o e q u a t i o n s ω {=a}_{max} {/v}_{max} = 4 w e k n o w t h a t, v_{max} = A×ω \Rightarrow A= \frac{v_{max}}{ω}, here A=amplitude \Rightarrow A = \frac{1}{4} = 0.25 m$

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