Question

A body performs linear S.H.M. having amplitude 5cm. When the displacement is 3 cm, find the ratio of its K.E. to P.E​

Answer 1

A = 5 cm = 0.05 m

T = 0.2 s

ω=2π/T=2π/0.2=10πrad/s

When displacement is y, then acceleration, a=−ω

2

y

Velocity, V=ω

r

2

−y

2

Case (a) When y=5cm=0.05m

a=−(10π)

2

×0.05=−5π

2

m/s

2

V=10π×

(0.05)

2

−(0.05)

2

=0

Case (b) When y=3cm=0.03m

a=−(10π)

2

×0.03=−3π

2

m/s

2

V=10π×

(0.05)

2

−(0.03)

2

=10π×0.04=0.4π m/s

Case (c) When y=0

a=−(10π)

2

×0=0

V=10π×

(0.05)

2

−0

2

=10π×0.05=0.5πm/s

Answer 2

Given: A body performs linear S.H.M. having amplitude 5 cm and the displacement is 3 cm.

To find: The ratio of its K.E. to P.E​..

Solution:

• When a body is performing linear simple harmonic motion, its kinetic energy is given by the formula,

        $K.E. = \frac{1}{2} m \omega ^{2} (A^{2} - x^{2})$

• Here, m is the mass of the body, ω is the angular frequency, A is the amplitude and x is the displacement of the body.

• The potential energy of the body is given by the formula,

        $P.E. = \frac{1}{2} m \omega ^{2} x^{2}$

• The ratio of the kinetic energy to the potential energy would be,

        $\frac{K.E.}{P.E.} = \frac{(A^{2} - x^{2})}{x^{2} }$

• In the given question, the amplitude is 5 cm and displacement is 3 cm.

        $\frac{K.E.}{P.E.} = \frac{(5^{2} - 3^{2})}{3^{2} }$

                $= \frac{16}{9}$

Therefore, the ratio of the K.E. to P.E​ of the body is 16/9.