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:
$ mg(h+y)=\frac{1}{2}xy^{2} $
$ y=\frac{mg}{k}+\frac{mg}{k}[ \sqrt{1+\frac{2hk}{mg}} ] $
So, amplitude of vibration is the maximum displacement from the equilibrium position
$ a=y-mg\text{/}k $
$ =\frac{mg}{k}[ \sqrt{1+\frac{2hk}{mg}} ] $
BETA
