Wave Mechanics Question 566
Question: A particle of mass 10 gm is describing S.H. M. along a straight line with period of 2 sec and amplitude of 10 cm. Its kinetic energy when it is at 5 cm from its equilibrium position is
Options:
A) $ 37.5{{\pi }^{2}},ergs $
B) $ 3.75{{\pi }^{2}},ergs $
C) $ 375{{\pi }^{2}},ergs $
D) $ 0.375{{\pi }^{2}},erg $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Kinetic energy $ K=\frac{1}{2}m{{\omega }^{2}}(a^{2}-y^{2}) $
$ =\frac{1}{2}\times 10\times {{( \frac{2\pi }{2} )}^{2}}[10^{2}-5^{2}]=375{{\pi }^{2}}ergs $
BETA
