Try Numerade free for 7 days Try Numerade free for 7 days Try Numerade free for 7 days Try Numerade free for 7 days Try Numerade free for 7 days Try Numerade free for 7 days Continue  Text Solution Solution : (i) As time period of oscillation os simple pendulum `T=2pisqrt((l)/(g))` is independent of amplitude of oscillation, hence time period of oscillation does no change with the change in the amplitude of oscillation. <br> (ii) Max. velocity, `v_(0)=omegaA`. When amplitude A is doubled, then max. velocity becomed double. <br> (iii) Max. acc., `a_(max)=omega^(2)A.` When amplitude A is doubled, then max, acceleration becomes double. <br> (iv) Total energy <br> `E=(1)/(2)m(2piv)^(2)A^(2)=2pi^(2)mv^(2)A^(2)` <br> When amplitude A is doubled, then the energy of oscillator becomes four times.
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What is the period of the system if the amplitude is doubled?
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