Question 175 of 194

A $1kg$ block attached to a spring vibrates with a frequency of $1Hz$ on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to a$8kg$block placed on the same table. So, the frequency of vibration of the $8kg$ block is

A$2Hz$B$\dfrac{1}{4}Hz$C$\dfrac{1}{2\sqrt{2}}Hz$D$\dfrac{1}{2}Hz$

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Answer: (D) $\dfrac{1}{2}Hz$

The frequency of $1kg$ block is $f=\dfrac{1}{2\pi }\sqrt{\dfrac{k}{m}}=1Hz$

$4{\pi }^{2}=\dfrac{k}{m},m=1$

Spring constant, $k=4{\pi }^{2}$

In parallel connection of springs, the equivalent spring constant is ${k}_{eq}=2k$.



$f=\dfrac{1}{2\pi }\sqrt{\dfrac{K\times 2}{8}}$

Frequency of $8kg$ block is$=\dfrac{1}{2}Hz$