Question

The maximury speed of particle
performing S.Hm is 0.08m/s.If max acceleration is 0.32 m/s^.calculate its
period and amplitude
​

Answer 1

Given :

➳ A particle is performing SHM.

• Max. velecity = 0.08m/s
• Max. acceleration = 0.32m/s²

To Find :

⟶ Time period and amplitude of particle.

SoluTion :

⇒ Max. velocity of particle performing SHM is given by

• v = Aω

⇒ Max. acceleration of particle performing SHM is given by

• a = Aω²

Where,

▪ A denotes amplitude

▪ ω denotes angular frequency

⏏ Taking ratio of max. acceleration and max. velocity, we get

➠ a/v = Aω²/Aω

➠ a/v = ω

➠ ω = 0.32/0.08

➠ ω = 4 $\longrightarrow$ (I)

➠ 2π/T = 4

➠ T = 2π/4

➠ T = π/2 s

Putting value of (I) in formula of maximum velocity, we get

➝ v = Aω

➝ 0.08 = A × 4

➝ A = 0.08/4

➝ A = 0.02m

◈ Time period = π/2 s

◈ Amplitude = 0.02 m