Question

A particle executes SHM on a straight line path.The amplitude of oscillation is 2 cm.when the displacement of particle from the mean position is 1cm,the numerical value of magnitude of acceleration is equal to the numerical value of magnitude of velocity. The frequency of SHM is

Answer 1

You can solve it further by putting value of √3 and π

Answer 2

Answer:

T = 2 π / √ 3

v_max =  2 √ 3 m / sec

a_max = - 6 m / sec²

Explanation:

Given :

Amplitude A = 2 cm

Displacement of particle from mean position x = 1 cm

It is said acceleration is equal to velocity of particle.

ω ( √ A² - x² ) =  ω² x

ω x = ( √ A² - x² )

Putting value here :

ω  = ( √ 4 - 1 )

ω = √ 3

We know :

ω = 2 π / T

T = 2 π / √ 3 .

Now for :

v_max = A ω

v_max = 2 × √ 3

v_max =  2 √ 3 m / sec

For a_max = - A ω²

a_max = - 2 ( √ 3 )²

a_max = - 6 m / sec²

Hence we get answer.