Work Energy And Power Question 331

Question: A body of mass m is released from a height h to a scale pan hung from a spring as shown in figure. The spring constant of the spring is k, the mass of the scale pan is negligible and the body does not bounce relative to the pan, then the amplitude of vibration is

Options:

A) $ \frac{mg}{k}\sqrt{( 1+\frac{2hk}{mg} )} $

B) $ \frac{mg}{k} $

C) $ \frac{mg}{k}[ 1+\sqrt{( 1+\frac{2hk}{mg} )} ] $

D) $ \frac{mg}{k}-\frac{mg}{k}[ \sqrt{( 1-\frac{2hk}{mg} )} ] $

Show Answer

Answer:

Correct Answer: A

Solution:

[a] $ mg(h+y)=1\text{/}2xy^{2} $

$ y=\frac{mg}{k}+\frac{mg}{k}[ \sqrt{1+\frac{2hk}{mg}} ] $

So, amplitude of vibration

$ a=y-mg\text{/}k $

$ =\frac{mg}{k}[ \sqrt{1+\frac{2hk}{mg}} ] $



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