Question

The equation of motion of a particle executing simple harmonic motion is given by $x = 3 \sin(6t+\frac{\pi}{6})$, where x is in metres and t is in seconds. The ratio of the potential and kinetic energies of the particle at time $t=0$ is

• A. $1:1$
• B. $1:4$
• C. $1:2$
• D. $1:3$

Hint:

Energies in SHM:

• $PE = \frac12m\omega^2x^2$, $KE = \frac12m\omega^2(A^2 - x^2)$.
• Energy ratio: $\fracPEKE = \fracx^2A^2 - x^2$.
• Use displacement at given $t$ to evaluate.

Answer 1

The Correct Option is D

The given SHM equation is $x = 3 \sin(6t + \frac{\pi}{6})$, so amplitude $A = 3$ m, and angular frequency $\omega = 6$ rad/s. At $t=0$, $x = 3 \sin(\frac{\pi}{6}) = 3 \times \frac{1}{2} = \frac{3}{2}$ m.
Potential energy: $PE = \frac{1}{2}m\omega^2x^2$. Kinetic energy: $KE = \frac{1}{2}m\omega^2(A^2 - x^2)$. So, $\frac{PE}{KE} = \frac{x^2}{A^2 - x^2}$.
Substituting: $\frac{(\frac{3}{2})^2}{3^2 - (\frac{3}{2})^2} = \frac{9/4}{9 - 9/4} = \frac{9}{27} = \frac{1}{3}$.
Hence, the ratio of PE to KE at $t=0$ is $1:3$.

Answer 2

Step 1: Identify the given equation of motion
The particle executes simple harmonic motion (SHM) with displacement:
\(x = 3 \sin\left(6t + \frac{\pi}{6}\right)\), where \(x\) is in meters and \(t\) in seconds.

Step 2: Recall energy expressions in SHM
Total mechanical energy \(E\) = constant = \(\frac{1}{2} m \omega^2 A^2\)
Potential energy (PE) at displacement \(x\):
\[ PE = \frac{1}{2} m \omega^2 x^2 \]
Kinetic energy (KE):
\[ KE = E - PE = \frac{1}{2} m \omega^2 (A^2 - x^2) \]

Step 3: Extract parameters
Amplitude \(A = 3\) m
Angular frequency \(\omega = 6\) rad/s
Time \(t = 0\)

Step 4: Calculate displacement \(x\) at \(t=0\)
\[ x = 3 \sin\left(0 + \frac{\pi}{6}\right) = 3 \times \frac{1}{2} = 1.5 \text{ m} \]

Step 5: Calculate ratio of PE to KE
\[ PE = \frac{1}{2} m \omega^2 x^2 \] \[ KE = \frac{1}{2} m \omega^2 (A^2 - x^2) \] Ratio:
\[ \frac{PE}{KE} = \frac{x^2}{A^2 - x^2} = \frac{(1.5)^2}{3^2 - (1.5)^2} = \frac{2.25}{9 - 2.25} = \frac{2.25}{6.75} = \frac{1}{3} \]

Step 6: Final answer
The ratio of potential energy to kinetic energy at \(t=0\) is \(1:3\).