Question

A body of 0.25 kg executes SHM given by m. Calculate the amplitude, angular
frequency, maximum velocity and maximum acceleration.

Answer 1

Answer:

The equation of motion of S.H.M x=0.5cos(100+

4

3π

). Comparing it with general equation motion of S.H.M, described as follows, we get, x=Acos(ωt+θ)

(1) Frequency=

2π

ω

, Where ω=100

=

2π

100

=(50π)Hz

(2)Initial phaseϕ=

4

3π

(3)Amplitude A=0.5cm

Maximum velocity=ωA=100×0.5=500m/s

(4)Maximum acceleration=ω

2

A=100

2

×0.5=50m/s

2

(5)Total energy=

2

1

mω

2

A

2

(Mass of the particle=0.1kg)

=

2

1

×0.1×(100)

2

×(0.5)

2

=125J