A particle undergoing simple harmonic motion has time-dependent displacement given by
x(t) = A sin(πt / 90).

The ratio of kinetic to potential energy of the particle at t = 210s will be:

A) 3
B) 1/9
C) 1/3
D) 1

Answer: C) 1/3
Explanation:
The displacement of the particle undergoing simple harmonic motion (SHM) is given by:
x(t)=Asin⁡(πt90)x(t) = A \sin\left(\dfrac{\pi t}{90}\right)
We are asked to find the ratio of kinetic energy (KE) to potential energy (PE) at t=210 st = 210 \, \text{s}.
Step-by-Step Solution:
1. Angular Frequency: The general equation for SHM is:
x(t)=Asin⁡(ωt)x(t) = A \sin(\omega t)
Here, comparing this with the given displacement equation, we have:
ω=π90\omega = \dfrac{\pi}{90}
2. Kinetic Energy (KE): The kinetic energy of a particle in SHM is given by:
KE=12mv2KE = \dfrac{1}{2} m v^2
where v=dxdtv = \dfrac{dx}{dt} is the velocity. To find v(t)v(t), differentiate the displacement equation:
v(t)=dxdt=A⋅π90cos⁡(πt90)v(t) = \dfrac{dx}{dt} = A \cdot \dfrac{\pi}{90} \cos\left( \dfrac{\pi t}{90} \right)
Therefore, the kinetic energy is:
KE=12m(A⋅π90cos⁡(πt90))2KE = \dfrac{1}{2} m \left( A \cdot \dfrac{\pi}{90} \cos\left( \dfrac{\pi t}{90} \right) \right)^2
3. Potential Energy (PE): The potential energy in SHM is given by:
PE=12kx2PE = \dfrac{1}{2} k x^2
where kk is the spring constant, and xx is the displacement. We know that for SHM:
k=mω2k = m \omega^2
Hence, the potential energy becomes:
PE=12mω2x2=12m(π90)2A2sin⁡2(πt90)PE = \dfrac{1}{2} m \omega^2 x^2 = \dfrac{1}{2} m \left( \dfrac{\pi}{90} \right)^2 A^2 \sin^2\left( \dfrac{\pi t}{90} \right)
4. Finding the Ratio of KE to PE: The total mechanical energy in SHM is the sum of kinetic and potential energy, and it is constant:
Etotal=KE+PE=12mω2A2E_{\text{total}} = KE + PE = \dfrac{1}{2} m \omega^2 A^2
The ratio of KE to PE at any time is:
KEPE=v2ω2x2\dfrac{KE}{PE} = \dfrac{v^2}{\omega^2 x^2}
Substituting the expressions for v(t)v(t) and x(t)x(t), we get:
KEPE=(A⋅π90cos⁡(πt90))2(π90)2A2sin⁡2(πt90)\dfrac{KE}{PE} = \dfrac{ \left( A \cdot \dfrac{\pi}{90} \cos\left( \dfrac{\pi t}{90} \right) \right)^2 }{ \left( \dfrac{\pi}{90} \right)^2 A^2 \sin^2\left( \dfrac{\pi t}{90} \right) }
Simplifying:
KEPE=cos⁡2(πt90)sin⁡2(πt90)\dfrac{KE}{PE} = \dfrac{\cos^2\left( \dfrac{\pi t}{90} \right)}{\sin^2\left( \dfrac{\pi t}{90} \right)}
Now, substitute t=210 st = 210 \, \text{s}:
πt90=π×21090=7π3\dfrac{\pi t}{90} = \dfrac{\pi \times 210}{90} = \dfrac{7\pi}{3}
This corresponds to an angle where:
cos⁡(7π3)=cos⁡(π3)=12\cos\left( \dfrac{7\pi}{3} \right) = \cos\left( \dfrac{\pi}{3} \right) = \dfrac{1}{2}
and
sin⁡(7π3)=sin⁡(π3)=32\sin\left( \dfrac{7\pi}{3} \right) = \sin\left( \dfrac{\pi}{3} \right) = \dfrac{\sqrt{3}}{2}
Substituting these values:
KEPE=(12)2(32)2=1434=13\dfrac{KE}{PE} = \dfrac{\left( \dfrac{1}{2} \right)^2}{\left( \dfrac{\sqrt{3}}{2} \right)^2} = \dfrac{\dfrac{1}{4}}{\dfrac{3}{4}} = \dfrac{1}{3}
Thus, the ratio of kinetic energy to potential energy at t=210 st = 210 \, \text{s} is 13\dfrac{1}{3}, which corresponds to option C.