Question

A particle of mass 5x10^-3 kg executes
S.H.M. and has an amplitude of 0.08 m.
Its frequency is 16 Hz. What is its
maximum velocity at the mean position *
O 30.23 m/s
O 8.038 m/s
O 1.25 m/s
O 8 m/s​

Answer 1

Given :

➳ Mass of particle = 0.005kg

➳ Amplitude (A) = 0.08m

➳ Frequency (f) = 16Hz

To Find :

➛ Velocity of particle at mean position (x = 0).

Solution :

➠ Velocity of particle executing SHM is given by

• v = ω √(A² - x²)

where,

◕ v denotes linear velocity

◕ ω denotes angular frequency

◕ A denotes amplitude

◕ x denotes distance from mean position

✴ Angular frequency of particle :

➝ ω = 2π × f

➝ ω = 2π × 16

➝ ω = 32π Hz

✴ Velocity at mean position :

⇒ v = ω √(A² - x²)

⇒ v = 32π × √[(0.08)² - (0)²]

⇒ v = 32π × √(0.08)²

⇒ v = 32 × 3.14 × 0.08

⇒ v = 8.038 m/s