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Mechanical Properties Of Fluids

Question
CBSEENPH11019197
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A particle executing simple harmonic motion has acceleration 5.12 cm/s2 when at 8 cm from the mean position. Find the time period of oscillation.

Solution

Using the formula of acceleration,
We have
                 a = straight omega squared straight y
So, time period of the oscillation is given by, 
therefore space space space space space space straight T space equals space 2 straight pi square root of 1 over straight omega squared end root equals 2 straight pi square root of straight y over straight a end root
Now, putting the values of acceleration and displacement give, we get
straight T equals 2 straight pi square root of fraction numerator 8 over denominator 5.12 end fraction end root equals 7.854 space straight s, is the time period of the oscillation. 

Some More Questions From Mechanical Properties of Fluids Chapter

What determines the natural frequency of an oscillator?

Is there any relation between uniform circular motion and simple harmonic motion?

What is periodic motion?

Write down the expression for the instantaneous velocity and acceleration of particle vibrating in simple harmonic motion.

How do you define the mean position of particle executing simple harmonic motion?

Define amplitude.

What is the net displacement of the particle executing simple harmonic motion in one complete oscillation?

Are all the periodic motions simple harmonic?

Does the time period of particle vibrating in simple harmonic motion depend upon its displacement?

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