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 (in Hz) of vibration of the 8 kg block is _____.

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Detailed Solution

If 1 kg block attached to a spring vibrates with a frequency,

$v = \frac{1}{2 π} \sqrt{\frac{k}{m}} = 1 Hz$

$\Rightarrow k = 4 π^{2} {Nm}^{- 1}$

When two springs are attached in parallel to an 8 kg block, then

$k_{eq} = k + k = 2 k$

Frequency, $v^{'} = \frac{1}{2 π} \sqrt{\frac{k_{eq}}{m^{'}}}$

$= \frac{1}{2 π} \sqrt{\frac{2 k}{8}} = \frac{1}{2 π} \sqrt{\frac{2 \times 4 π^{2}}{8}} = \frac{1}{2} Hz = 0 .5 Hz$

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