Question

The acceleration of the particle performing S.H.M. is 0.168 m/s2, when the particle is at a distance of 2.2 cm from its extreme position. The amplitude of oscillation, if maximum acceleration of the particle is 0.3 m/s2, is​

Answer 1

Answer:

3.9cm

Explanation:

we know that,

a=0.168 m/s²

a(max)=0.3m/s²

let us considered a position from extreme position as x=2.2cm=2.2* 10-2m

therefore we can write here and equation as

x=A coswt .......1

double differentiation gives

a=Aw² coswt .......2

from equation 1

2.2*10-2/A = coswt

putting this value in equation 2

0.168=Aw² (2.2*10-2/A)

w²=7.63 ......3

putting the value of equation 3

a(max)=Aw²

0.3=A*7.63

A= 0.039m or 3.9cm