Question

A particle performing linear S.H.M. has
maximum velocity of 20 cm/s and maximum
acceleration of 80 cm/s2. Find the amplitude and
period of oscillation.​

Answer 1

Explanation:

We know,

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

where,

a = amplitude

w = angular frequency

α = phase difference

according to this question

Aw^2 = 80 cm/s^2 ............... ( 1 )

Aw = 20 cm/s ........................ ( 2 )

Now dividing equation ( 1 ) and ( 2 ) we get,

w = 4 s^-1

w = 2π / T

So, T = 1.57 sec.

Now, put the value of 'w' in equation ( 2 ) we get,

A = 5 cm.

= 5 × 10^-2 m.