Question:
An object mass \(2\,\text{kg}\) is attached to a spring with spring constant \(8\,\text{N m}^{-1}\). If the object is executing simple harmonic motion then the number of cycles it completes in \(66\,\text{s}\) is
An object mass \(2\,\text{kg}\) is attached to a spring with spring constant \(8\,\text{N m}^{-1}\). If the object is executing simple harmonic motion then the number of cycles it completes in \(66\,\text{s}\) is
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For a mass-spring SHM system,
\[
\omega=\sqrt{\frac{k}{m}}
\]
and
\[
T=2\pi\sqrt{\frac{m}{k}}.
\]
Number of oscillations \(=\dfrac{\text{Total Time}}{T}\).
Updated On: Jun 18, 2026
- \(21\)
- \(16\)
- \(28\)
- \(12\)
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The Correct Option is A
Solution and Explanation
Step 1: Calculate the angular frequency.
For a mass-spring system, \[ \omega=\sqrt{\frac{k}{m}}. \] Given, \[ k=8\,\text{N m}^{-1} \] and \[ m=2\,\text{kg}. \] Therefore, \[ \omega=\sqrt{\frac{8}{2}} \] \[ \omega=\sqrt{4} \] \[ \omega=2\,\text{rad s}^{-1}. \]
Step 2: Calculate the time period.
The time period is \[ T=\frac{2\pi}{\omega}. \] Substituting \(\omega=2\), \[ T=\frac{2\pi}{2} \] \[ T=\pi\ \text{s}. \] Using \[ \pi \approx \frac{22}{7}, \] \[ T=\frac{22}{7}\ \text{s}. \]
Step 3: Find the number of oscillations in \(66\,\text{s}\).
Number of cycles, \[ n=\frac{\text{Total Time}}{\text{Time Period}} \] \[ n=\frac{66}{22/7} \] \[ n=66\times \frac{7}{22} \] \[ n=3\times 7 \] \[ n=21. \]
Step 4: Final conclusion.
Hence, the number of cycles completed in \(66\,\text{s}\) is \[ \boxed{21} \]
For a mass-spring system, \[ \omega=\sqrt{\frac{k}{m}}. \] Given, \[ k=8\,\text{N m}^{-1} \] and \[ m=2\,\text{kg}. \] Therefore, \[ \omega=\sqrt{\frac{8}{2}} \] \[ \omega=\sqrt{4} \] \[ \omega=2\,\text{rad s}^{-1}. \]
Step 2: Calculate the time period.
The time period is \[ T=\frac{2\pi}{\omega}. \] Substituting \(\omega=2\), \[ T=\frac{2\pi}{2} \] \[ T=\pi\ \text{s}. \] Using \[ \pi \approx \frac{22}{7}, \] \[ T=\frac{22}{7}\ \text{s}. \]
Step 3: Find the number of oscillations in \(66\,\text{s}\).
Number of cycles, \[ n=\frac{\text{Total Time}}{\text{Time Period}} \] \[ n=\frac{66}{22/7} \] \[ n=66\times \frac{7}{22} \] \[ n=3\times 7 \] \[ n=21. \]
Step 4: Final conclusion.
Hence, the number of cycles completed in \(66\,\text{s}\) is \[ \boxed{21} \]
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