Question

A particle is doing simple harmonic motion of amplitude 0.06 m and time period 3.14 s. The maximum velocity of the particle is _______ cm/s.

Answer 1

Correct Answer: 12

We know:

\[ v_{\text{max}} = \omega A \quad \text{at mean position} \]

\[ \omega = \frac{2\pi}{T} \quad \text{and} \quad v_{\text{max}} = \frac{2\pi}{T} \times A \]

\[ v_{\text{max}} = \frac{2\pi}{3.14} \times 0.06 = 0.12 \, \text{m/s} \]

\[ v_{\text{max}} = 12 \, \text{cm/s} \]

Answer 2

This problem asks for the maximum velocity of a particle undergoing Simple Harmonic Motion (SHM). We are provided with the amplitude and the time period of the motion. The final answer needs to be expressed in cm/s.

Concept Used:

For a particle executing Simple Harmonic Motion, its velocity \(v\) at any displacement \(x\) from the mean position is given by:

\[ v = \omega \sqrt{A^2 - x^2} \]

where \(A\) is the amplitude and \(\omega\) is the angular frequency. The velocity is maximum when the particle is at its mean position, i.e., when \(x = 0\). The formula for the maximum velocity (\(v_{\text{max}}\)) is therefore:

\[ v_{\text{max}} = A\omega \]

The angular frequency \(\omega\) is related to the time period \(T\) by the formula:

\[ \omega = \frac{2\pi}{T} \]

Step-by-Step Solution:

Step 1: Identify the given values and convert them to the required units.

Amplitude, \(A = 0.06 \text{ m}\).

Since the final answer is required in cm/s, we convert the amplitude to cm:

\[ A = 0.06 \text{ m} \times 100 \frac{\text{cm}}{\text{m}} = 6 \text{ cm} \]

Time period, \(T = 3.14 \text{ s}\).

Step 2: Calculate the angular frequency (\(\omega\)).

The value of the time period is given as 3.14 s, which is a good approximation for \(\pi\). So, we can take \(T \approx \pi\) s.

\[ \omega = \frac{2\pi}{T} \]

Substituting the value of T:

\[ \omega = \frac{2\pi}{3.14 \text{ s}} \approx \frac{2\pi}{\pi \text{ s}} = 2 \text{ rad/s} \]

Step 3: Calculate the maximum velocity (\(v_{\text{max}}\)) using the formula \(v_{\text{max}} = A\omega\).

Substitute the values of amplitude \(A\) (in cm) and angular frequency \(\omega\):

\[ v_{\text{max}} = (6 \text{ cm}) \times (2 \text{ rad/s}) \]

Final Computation & Result:

Performing the multiplication gives the maximum velocity:

\[ v_{\text{max}} = 12 \text{ cm/s} \]

The maximum velocity of the particle is 12 cm/s.