A $1 \,kg$ block attached to a spring vibrates with a frequency of $1\, Hz$ on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an $8\, kg$ block placed on the same table. So, the frequency of vibration of the $8\, kg$ block is ..... $Hz$
JEE MAIN 2017, Medium
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Frequency of spring $(f)=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}=1 \mathrm{Hz}$
$\Rightarrow 4 \pi^{2}=\frac{\mathrm{k}}{\mathrm{m}}$
If block of mass $\mathrm{m}=1 \mathrm{kg}$ is attached then, $k=4 \pi^{2}$
Now, identical springs are attached in parallel with mass
$\mathrm{m}=8 \mathrm{kg}$, Hence, $k_{e q}=2 k$
$f^{\prime}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k} \times 2}{8}}=\frac{1}{2} \mathrm{Hz}$
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