Question:
A particle is executing SHM of amplitude \(4\text{ cm}\) and time period \(4\text{ s}\). The time taken by it to move from positive extreme position to half the amplitude is
A particle is executing SHM of amplitude \(4\text{ cm}\) and time period \(4\text{ s}\). The time taken by it to move from positive extreme position to half the amplitude is
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Standard SHM time intervals to memorize:
• Mean position (\(0\)) to half amplitude (\(A/2\)): \(t = T/12\)
• Half amplitude (\(A/2\)) to extreme (\(A\)): \(t = T/6\)
• Extreme (\(A\)) to half amplitude (\(A/2\)): \(t = T/6\) Here, the particle moves from \(A\) to \(A/2\), so \(t = T/6 = 4/6 = 2/3\text{ s}\).
• Mean position (\(0\)) to half amplitude (\(A/2\)): \(t = T/12\)
• Half amplitude (\(A/2\)) to extreme (\(A\)): \(t = T/6\)
• Extreme (\(A\)) to half amplitude (\(A/2\)): \(t = T/6\) Here, the particle moves from \(A\) to \(A/2\), so \(t = T/6 = 4/6 = 2/3\text{ s}\).
Updated On: May 13, 2026
- \(1\text{ s}\)
- \(\frac{1}{3}\text{ s}\)
- \(\frac{2}{3}\text{ s}\)
- \(\sqrt{\frac{2}{3}}\text{ s}\)
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The Correct Option is C
Solution and Explanation
Concept:
For a particle executing Simple Harmonic Motion (SHM) starting from the positive extreme position, its displacement \(x\) at any time \(t\) is given by the cosine function:
\[
x = A \cos(\omega t)
\]
where \(A\) is the amplitude and \(\omega\) is the angular frequency.
Step 1: Identifying given parameters and calculating angular frequency.
Given:
• Amplitude, \(A = 4\text{ cm}\)
• Time period, \(T = 4\text{ s}\) Angular frequency \(\omega\) is: \[ \omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2} \text{ rad/s} \] The target position is half the amplitude: \[ x = \frac{A}{2} = \frac{4}{2} = 2\text{ cm} \]
Step 2: Applying the displacement equation.
Since the particle starts from the positive extreme, we use: \[ x = A \cos(\omega t) \] Substitute the values of \(x\), \(A\), and \(\omega\): \[ 2 = 4 \cos\left(\frac{\pi}{2} t\right) \] \[ \cos\left(\frac{\pi}{2} t\right) = \frac{2}{4} = \frac{1}{2} \]
Step 3: Solving for time \(t\).
We know that \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\). Therefore: \[ \frac{\pi}{2} t = \frac{\pi}{3} \] Solving for \(t\): \[ t = \frac{\pi}{3} \times \frac{2}{\pi} \] \[ t = \frac{2}{3} \text{ s} \]
Step 4: Selecting the correct option.
From the calculation above, the time taken is: \[ \boxed{\frac{2}{3}\text{ s}} \] Therefore, the correct option is: \[ \boxed{\text{Option (C)}} \]
Step 1: Identifying given parameters and calculating angular frequency.
Given:
• Amplitude, \(A = 4\text{ cm}\)
• Time period, \(T = 4\text{ s}\) Angular frequency \(\omega\) is: \[ \omega = \frac{2\pi}{T} = \frac{2\pi}{4} = \frac{\pi}{2} \text{ rad/s} \] The target position is half the amplitude: \[ x = \frac{A}{2} = \frac{4}{2} = 2\text{ cm} \]
Step 2: Applying the displacement equation.
Since the particle starts from the positive extreme, we use: \[ x = A \cos(\omega t) \] Substitute the values of \(x\), \(A\), and \(\omega\): \[ 2 = 4 \cos\left(\frac{\pi}{2} t\right) \] \[ \cos\left(\frac{\pi}{2} t\right) = \frac{2}{4} = \frac{1}{2} \]
Step 3: Solving for time \(t\).
We know that \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\). Therefore: \[ \frac{\pi}{2} t = \frac{\pi}{3} \] Solving for \(t\): \[ t = \frac{\pi}{3} \times \frac{2}{\pi} \] \[ t = \frac{2}{3} \text{ s} \]
Step 4: Selecting the correct option.
From the calculation above, the time taken is: \[ \boxed{\frac{2}{3}\text{ s}} \] Therefore, the correct option is: \[ \boxed{\text{Option (C)}} \]
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