Wave Mechanics Question 548
Question: A body executing linear simple harmonic motion has a velocity of 3 m/s when its displacement is 4 cm and a velocity of 4 m/s when its displacement is 3 cm. What is the amplitude of oscillation?
Options:
A) 5 cm
B) 7.5 cm
C) 10 cm
D) 12.5 cm
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Velocity in SHM is given by $ v=\omega \sqrt{a^{2}-y^{2}} $ At $ y=4cm=0.04m,,v=3m/s $
$ \
Therefore 3=\omega \sqrt{a^{2}-{{(0.04)}^{2}}} $ ?. (i) At $ y=3cm=0.03m,,v=4m/s $
$ \
Therefore 4=\omega \sqrt{a^{2}-{{(0.03)}^{2}}} $ ?. (ii) Dividing (2) by (1), we get a =0.05=5cm
BETA
