Wave Mechanics Question 554
Question: The amplitude of a particle executing SHM is 4 cm. At the mean position the speed of the particle is 16 cm/sec. The distance of the particle from the mean position at which the speed of the particle becomes $ 8\sqrt{3},cm/s $ , will be
Options:
A) $ 2\sqrt{3,}cm $
B) $ \sqrt{3,}cm $
C) 1cm
D) 2cm
Show Answer
Answer:
Correct Answer: D
Solution:
[d] At mean position velocity is maximum i.e., $ {v _{\max }}=\omega a\Rightarrow \omega \frac{{v _{\max }}}{a}=\frac{16}{4}=4 $
$ \
Therefore v=\omega \sqrt{a^{2}-y^{2}}\Rightarrow 8\sqrt{3}=4\sqrt{4^{2}-y^{2}} $
$ \Rightarrow ,192=16(16-y^{2})\Rightarrow 12=16-y^{2} $
$ \Rightarrow ,y=2cm. $
BETA
