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 $1.0 m s^{- 1}$ . What is the period of oscillation?

0.1 s

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0.2 s

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Solution

The correct option is D
Displacement x = +0.01 m. Therefore, acceleration $a = - 1.0 m s^{- 2} .$
Now a = $- ω^{2} x$
Or -1.0 = $- ω^{2} \times 0.01$
Which gives, $ω = 10 r a d s^{- 1}$ . Therefore, $T = \frac{2 π}{ω} = \frac{π}{5} s .$
Hence the correct choice is (d).

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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?

The correct option is D Displacement x = +0.01 m. Therefore, acceleration a=−1.0ms−2. Now a = −ω2x Or -1.0 =−ω2×0.01 Which gives, ω=10 rad s−1 . Therefore, T=2πω=π5s. Hence the correct choice is (d).

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 $1.0 m s^{- 1}$ . What is the period of oscillation?

The correct option is D
Displacement x = +0.01 m. Therefore, acceleration $a = - 1.0 m s^{- 2} .$
Now a = $- ω^{2} x$
Or -1.0 = $- ω^{2} \times 0.01$
Which gives, $ω = 10 r a d s^{- 1}$ . Therefore, $T = \frac{2 π}{ω} = \frac{π}{5} s .$
Hence the correct choice is (d).