Question

An object of mass 0.5 kg is executing simple harmonic motion. Its amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant t = T/4 starting from mean position. Assume that the initial phase of the oscillation is zero.

• A. $6.2 \times 10^{-3}$ J
• B. $1.2 \times 10^3$ J
• C. 0.62 J
• D. $6.2 \times 10^3$ J

Hint:

For an object in SHM starting from the mean position, at $t=T/4$ it reaches the extreme position where its velocity is zero and its potential energy is maximum. At this point, the entire mechanical energy is potential energy, $E_{total} = P.E._{max} = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2A^2$.

Answer 1

The Correct Option is C

The formula for the potential energy (P.E.) of an object in Simple Harmonic Motion (SHM) is:
$P.E. = \frac{1}{2} m \omega^2 x^2$
Where m is mass, $\omega$ is angular frequency, and x is the displacement from the mean position.
The equation for displacement in SHM starting from the mean position (initial phase zero) is:
$x(t) = A \sin(\omega t)$
First, let's find the angular frequency $\omega$.
$\omega = \frac{2\pi}{T} = \frac{2\pi}{0.2} = 10\pi$ rad/s.
Next, find the displacement x at the given time instant $t = T/4$.
$x(T/4) = A \sin\left(\omega \frac{T}{4}\right) = A \sin\left(\frac{2\pi}{T} \frac{T}{4}\right) = A \sin\left(\frac{\pi}{2}\right) = A$.
So, at $t=T/4$, the object is at its extreme position (amplitude).
Convert amplitude to SI units: $A = 5 \text{ cm} = 0.05 \text{ m}$.
So, $x = 0.05 \text{ m}$.
Now, calculate the potential energy at this position.
$P.E. = \frac{1}{2} m \omega^2 A^2$.
$P.E. = \frac{1}{2} \times (0.5 \text{ kg}) \times (10\pi \text{ rad/s})^2 \times (0.05 \text{ m})^2$.
$P.E. = \frac{1}{2} \times 0.5 \times 100\pi^2 \times 0.0025$.
$P.E. = 0.25 \times 100\pi^2 \times 0.0025 = 25\pi^2 \times 0.0025 = 0.0625 \pi^2 \text{ J}$.
Using the approximation $\pi^2 \approx 9.87$:
$P.E. \approx 0.0625 \times 9.87 \approx 0.616875 \text{ J}$.
This value is approximately 0.62 J.