Question

A mass suspended from one end of a helical spring undergoes vertical simple harmonic motion

with an amplitude of 2.0 cm. if three complete oscillations are made in 4.0 s, what is the

acceleration of the mass at
(a) the equilibrium position,
(b) the position of maximum

displacement?

Answers
(a) 0
(b) 0.44 m/s²​

Answer 1

Answer :

Amplitude of SHM = 2 cm = 0.02m

No. of oscillations made in 4s = 3

We have to find acceleration of mass at

• Equilibrium position
• Maximum position

Let's solve it step by step!

★ Step - 1 [Time period of oscillation]

We know that, The time required by a body to complete one oscillation is known as the time period$.$

➠ T = t / n

➠ T = 4/3

➠ T = 1.33 s

★ Step - 2 [Angular frequency]

Angular frequency, ω = 2π/T

➠ ω = 2 × 3.14/1.33

➠ ω = 4.72 rad/s

★ Step - 3 [Acceleration]

Acceleration of particle performing SHM is given by a = ω²x

• a denotes acceleration
• ω denotes angular frequency
• x denotes distance from mean position

(1) Acceleration at mean position :

➙ a = ω²x

at mean position x = 0

∴ a = 0 m/s²

(2) Acceleration at extreme position :

➙ a' = ω²x

at extreme position x = A

➙ a' = (4.72)² × 0.02

➙ a' = 22.27 × 0.02

∴ a' = 0.44 m/s²