A particle of mass 200 g executes linear simple harmonic motion with an amplitude 10 cm. When the particles at a point midway between the mean and the extreme position, its kinetic energy is 3π2×10−3J. Assuming the initial phase to be 2π3, the equation of motion of the particle will be :

y=10 sin (2πt+2π3)cm

y=10 sin (4πt+2π3)cm

y=10 cos (2πt+π6)cm

y=10 cos (2πt+π3)cm

A

y=10 sin (2πt+2π3)cm

B

y=10 cos (2πt+π6)cm

C

y=10 cos (2πt+π3)cm

D

y=10 sin (4πt+2π3)cm

Open in App

Solution

Verified by Toppr

Was this answer helpful?

0

Similar Questions

Q1

A particle of mass 200 g executes linear simple harmonic motion with an amplitude 10 cm. When the particles at a point midway between the mean and the extreme position, its kinetic energy is 3π2×10−3J. Assuming the initial phase to be 2π3, the equation of motion of the particle will be :

View Solution

Q2

A particle of mass 100g is executing SHM with amplitude of 10 cm. When the particle passes through the mean position at t=0 its kinetic energy is 8 mJ. What is the equation of simple harmonic motion if initial phase is zero?

View Solution

Q3

A particle of mass 400g is executing SHM of amplitude 0.4m. When it passes through the mean position, its kinetic energy is 32×10−3J. If the initial phase of oscillation is π4, then the equation of motion of the particle is

View Solution

Q4

A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is:

View Solution

Q5

A point particle of mass 0.1kg is executing SHM with amplitude of 0.1m. When the particle passes through the mean position, its kinetic energy is 8×10−3J. If the initial phase of oscillation is 45∘, the equation of motion of this particle is