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An object of mass $2 \mathrm{~kg}$ is attached to a spring with spring constant $8 \mathrm{Nm}^{-1}$. If the object is executing simple harmonic motion then the number of cycles it completes in $66 \mathrm{~s}$ is
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The correct answer is:
$21$
Time period of a spring mass system is
$T=2 \pi \sqrt{\frac{m}{k}}$
Here, $\quad m=2 \mathrm{~kg}$ and $k=8 \mathrm{~N} / \mathrm{m}$
So, time for 1 oscillation,
$T=2 \pi \sqrt{\frac{2}{8}}=2 \times \pi \times \frac{1}{2}$
or $\quad T=\pi \mathrm{s}$
Number of cycles in 66 seconds is
$n=\frac{t}{T}=\frac{66}{\pi}$
$\Rightarrow \quad n=\frac{66 \times 7}{22}=21$
Hence, object complete 21 oscillations in given time.
$T=2 \pi \sqrt{\frac{m}{k}}$
Here, $\quad m=2 \mathrm{~kg}$ and $k=8 \mathrm{~N} / \mathrm{m}$
So, time for 1 oscillation,
$T=2 \pi \sqrt{\frac{2}{8}}=2 \times \pi \times \frac{1}{2}$
or $\quad T=\pi \mathrm{s}$
Number of cycles in 66 seconds is
$n=\frac{t}{T}=\frac{66}{\pi}$
$\Rightarrow \quad n=\frac{66 \times 7}{22}=21$
Hence, object complete 21 oscillations in given time.
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