Question
A body oscillates simple harmonically with an amplitude of 0.05 m. At a certain instant of time its displacement is 0.01 m and acceleration is 1.0ms−1 . What is the period of oscillation?
Answer: Option D
:
D
Displacement x = +0.01 m. Therefore, acceleration a=−1.0ms−2.
Now a = −ω2x
Or -1.0 =−ω2×0.01
Which gives, ω=10rads−1 . Therefore, T=2πω=π5s.
Hence the correct choice is (d).
Was this answer helpful ?
:
D
Displacement x = +0.01 m. Therefore, acceleration a=−1.0ms−2.
Now a = −ω2x
Or -1.0 =−ω2×0.01
Which gives, ω=10rads−1 . Therefore, T=2πω=π5s.
Hence the correct choice is (d).
Was this answer helpful ?
More Questions on This Topic :
Latest Videos
Latest Test Papers
Submit Solution