Question

The displacement of a particle executing SHM is given by \( x = 10 \sin\left(\omega t + \frac{\pi}{3}\right) \, \text{m} \). The time period of motion is \( 3.14 \, \text{s} \). The velocity of the particle at \( t = 0 \) is ______ m/s.

Answer 1

Correct Answer: 10

Given:
\[ T = 3.14 = \frac{2\pi}{\omega}. \]

Solving for \( \omega \):
\[ \omega = 2 \, \text{rad/s}. \] 

The displacement \( x \) is given by:
\[ x = 10 \sin\left(\omega t + \frac{\pi}{3}\right). \] 

To find the velocity \( v \), differentiate \( x \) with respect to \( t \):
\[ v = \frac{dx}{dt} = 10\omega \cos\left(\omega t + \frac{\pi}{3}\right). \] 

At \( t = 0 \):
\[ v = 10\omega \cos\left(\frac{\pi}{3}\right) = 10 \times 2 \times \frac{1}{2} = 10 \, \text{m/s}. \] 

Answer: 10 m/s

Answer 2

Step 1: Write the given equation of motion
The displacement of the particle in simple harmonic motion (SHM) is given as:
\[ x = 10 \sin\left(\omega t + \frac{\pi}{3}\right) \] Here, the amplitude \(A = 10\,\text{m}\) and the time period \(T = 3.14\,\text{s}\).

Step 2: Find the angular frequency
The angular frequency \(\omega\) is related to the time period by the formula:
\[ \omega = \frac{2\pi}{T} \] Substitute \(T = 3.14\,\text{s}\):
\[ \omega = \frac{2\pi}{3.14} \approx 2\,\text{rad/s} \]

Step 3: Expression for velocity in SHM
The velocity of a particle in SHM is given by the derivative of displacement with respect to time:
\[ v = \frac{dx}{dt} = A\omega \cos\left(\omega t + \frac{\pi}{3}\right) \]

Step 4: Substitute the given values at \(t = 0\)
At \(t = 0\):
\[ v = A\omega \cos\left(\frac{\pi}{3}\right) \] Substitute \(A = 10\) and \(\omega = 2\):
\[ v = 10 \times 2 \times \cos\left(\frac{\pi}{3}\right) \] \[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \] Hence,
\[ v = 20 \times \frac{1}{2} = 10\,\text{m/s} \]

Step 5: Interpretation
At \(t = 0\), the velocity of the particle is \(10\,\text{m/s}\). The positive sign indicates that the particle is moving in the positive direction of the x-axis at that instant.

Final Answer:

10