A particle is executing simple harmonic motion in one dimension. If the amplitude of oscillations is 0.2 cm and if its velocity at the mean position is 5 m / s, then the angular frequency of the oscillation is

Question

A particle is executing simple harmonic motion in one dimension. If the amplitude of oscillations is 0.2 cm and if its velocity at the mean position is 5 m / s, then the angular frequency of the oscillation is (A) 1000 rad / s (B) 1500 rad / s (C) 2000 rad / s (D) 2500 rad / s

Tardigrade Admin · Accepted Answer

Given, amplitude, A=0.2 cm =2 × 10-3 m Velocity at mean position in SHM v max =5 m / s We know that, v max =A ω ⇒ ω=(v max /A) =(5/2 × 10-3)=25 × 103=2500 rads -1

Q. A particle is executing simple harmonic motion in one dimension. If the amplitude of oscillations is $0.2 c m$ and if its velocity at the mean position is $5 m / s$ , then the angular frequency of the oscillation is

1000 rad / s

1500 rad / s

2000 rad / s

2500 rad / s

Given, amplitude, $A = 0.2 c m = 2 \times 1 0^{- 3} m$
Velocity at mean position in SHM
$v_{m a x} = 5 m / s$
We know that,
$v_{m a x} = A ω$
$\Rightarrow ω = \frac{v _{m a x}}{A}$
$= \frac{5}{2 \times 1 0 ^{- 3}} = 25 \times 1 0^{3} = 2500 r a d s^{- 1}$

Q. A particle is executing simple harmonic motion in one dimension. If the amplitude of oscillations is 0.2cm and if its velocity at the mean position is 5m/s, then the angular frequency of the oscillation is