Question

a particle executing SHM has a maximum velocity of 40m/s and a maximum acceleration of 50m/s ^2 .find it's amplitude and time period​

Answer 1

Explanation:

$x = \: a \sin(wt \: + \: \alpha )$

where,

a = amplitude

w = angular frequency

α = phase difference

Aw^2 = 50 m/s^2 ..................( 1 )

Aw = 40 m/s ....................( 2 )

dividing equation ( 1 ) and ( 2 ) We get,

w = 5 ÷ 4

w= 2π ÷ T

T = 5.02 sec.

Put the value of 'w' in equation ( 2 ), We get

A = 32m.