The displacement of a particle in simple harmonic motion (SHM) is given by y=√(3π)sin100/(π)t+(π)/4. What will be the displacement of the particle…

Question

The displacement of a particle in simple harmonic motion (SHM) is given by y=√(3π)sin100/(π)t+(π)/4. What will be the displacement of the particle from the mean position when its kinetic energy is eight times that of its potential energy?

MARKS Content Team · Accepted Answer

Correct answer: (A) √(π3). The formula of kinetic energy and potential energy in S.H.M. K.E.=1/2mω^2A^2-y^2 & P.E.=1/2mω^2y^2 , in the question given that, K.E.=8P.E.⇒1/2mω^2A^2-y^2=8×1/2mω^2y^2⇒A^2-y^2=8y^2⇒9y^2=A^2⇒y=(±A)/3 Now in the above equation, y=√(3π)sin100/(π)t+(π)/4 the value of amplitude, A=√(3π), ⇒y=A/3=√(3π)3=√(π)√3

The displacement of a particle in simple harmonic motion (SHM) is given by $y = \sqrt{3 π} \sin (\frac{100}{π} t + \frac{π}{4}) .$ What will be the displacement of the particle from the mean position when its kinetic energy is eight times that of its potential energy?

Solution

Relevant Questions

The displacement of a particle in simple harmonic motion (SHM) is given by y=3πsin100πt+π4. What will be the displacement of the particle from the mean position when its kinetic energy is eight times that of its potential energy?

Solution

Relevant Questions

The displacement of a particle in simple harmonic motion (SHM) is given by $y = \sqrt{3 π} \sin (\frac{100}{π} t + \frac{π}{4}) .$ What will be the displacement of the particle from the mean position when its kinetic energy is eight times that of its potential energy?