Question

a body of mass m tied to a spring performs shm with period 2 seconds if the mass is increased by 3 m, what will be the period of SHM​

Answer 1

Given info : A body of mass m tied to a spring performs simple harmonic motion with period 2 seconds. if mass is increased by 3m.

To find : the period of simple harmonic motion is..

solution : time period of simple harmonic is given by, T = 2π√{m/K}

where K is spring constant, m is mass of body performing SHM.

here T = 2 seconds

so, 2 = 2π√{m/K}

⇒1/π² = m/K. ....(1)

now mass is increased by 3m

so, new mass of body is , m' = (m + 3m) = 4m

T' = 2π√{m'/K}

= 2π√{4m/K}

from equation (1),

= 2π√{4 × 1/π²}

= 2π × 2/π

= 4 seconds

Therefore the time period of SHM will be 4 seconds.