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.
Correct Answer: 12
Approach Solution - 1
We know:
\[ v_{\text{max}} = \omega A \quad \text{at mean position} \]
\[ \omega = \frac{2\pi}{T} \quad \text{and} \quad v_{\text{max}} = \frac{2\pi}{T} \times A \]
\[ v_{\text{max}} = \frac{2\pi}{3.14} \times 0.06 = 0.12 \, \text{m/s} \]
\[ v_{\text{max}} = 12 \, \text{cm/s} \]
Approach Solution -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.
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