Question

A particle of mass 200 g executes SHM. The restoring force is provided by spring of spring constant 80 Nm-7. What is magnitude of acceleration when mass is 2 cm away from equilibrium position?​

Answer 1

Given info : A particle of mass 200 g executes SHM. The restoring force is provided by spring of spring constant 80 N/m.

To find : the magnitude of acceleration when mass is 2cm away from equilibrium position is ..

solution : at equilibrium,

spring force = mass × acceleration

⇒ Kx = ma

here, spring constant, K = 80 N/m

elongation of spring from equilibrium, x = 2cm = 0.02m

mass of particle executes shm, m = 200g = 0.2 kg

so, 80 × 0.02 = 0.2 × a

⇒ a = 8 m/s²

therefore the magnitude of acceleration of mass is 8 m/s²

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