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Question: A particle of mass \(0.1\,kg\) is executing SHM of amplitude \(0.1\,m\) , When the particle passes t...

A particle of mass 0.1kg0.1\,kg is executing SHM of amplitude 0.1m0.1\,m , When the particle passes through mean position passes through mean position K.E = 8×103J8 \times {10^{ - 3}}\,J .Find equation of particle if [ϕ=π3]\left[ {\phi = \dfrac{\pi }{3}} \right].

Explanation

Solution

This problem is based on Simple harmonic motion. It is required to find the angular frequency of the particle and angular frequency is used to find the equation of the displacement for a particle which executing simple harmonic motion

Formula used:
y=Asin(ωt+ϕ)y = A\sin \left( {\omega t + \phi } \right)
Where, y=y = Displacement, A=A = Amplitude, ω=\omega = Angular frequency and ϕ=\phi = Initial phase angle.

Complete step by step answer:
Given, Mass (m)=0.1kg\left( m \right) = 0.1kg.
Amplitude (A)=0.1m\left( A \right) = 0.1m
Kinetic energy (K.E)=8×103J\left( {K.E} \right) = 8 \times {10^{ - 3}}J
Initial phase of oscillation (ϕ)=π3\left( \phi \right) = \dfrac{\pi }{3}
We know that, kinetic energy of particle is given as
Kinetic energy (K.E)=12mv2\left( {K.E} \right) = \dfrac{1}{2}m{v^2} ………(1)\left( 1 \right)
And also velocity at mean position is given by
Velocity (v)=ωA\left( v \right) = \omega A……….(2)\left( 2 \right)
Where ω=\omega = angular frequency and A=A = Amplitude.
Substituting equation (2)\left( 2 \right) in equation (1)\left( 1 \right) the expression for kinetic energy becomes
K.E=12m(ωA)2K.E = \dfrac{1}{2}m{\left( {\omega A} \right)^2} ……….(3)\left( 3 \right)
Substituting the given data in equation (3)\left( 3 \right) we get,
8×103=12×0.1×ω2×(0.1)28 \times {10^{ - 3}} = \dfrac{1}{2} \times 0.1 \times {\omega ^2} \times {\left( {0.1} \right)^2}
On simplifying the above equation we get
ω2=16{\omega ^2} = 16
ω=4\Rightarrow \omega = 4
The general equation for displacement (y)\left( y \right) for simple harmonic motion is given as,
y=Asin(ωt+ϕ)y = A\sin \left( {\omega t + \phi } \right) ……….(4)\left( 4 \right)
Substituting the values we get displacement as
y=0.1sin(4t+π3)\therefore y = 0.1\sin \left( {4t + \dfrac{\pi }{3}} \right)

Note: For the problems similar to given where mean position is mentioned, we must remember the conditions for the initial position and mean position. At the initial position, displacement, acceleration and potential energy will be maximum and velocity, kinetic energy will be minimum. Whereas at the mean position, velocity and kinetic energy will be maximum and displacement, acceleration and potential energy is zero. Velocity will always be inverse then displacement and acceleration because displacement and acceleration follow the sine curve, while velocity follows the cosine curve.