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The periodic time of a simple pendulum is $T_1$. Now if the point of suspension of this pendulum starts moving along the vertical direction according to the equation $y = kt^2$, the periodic time of the pendulum becomes $T_2$ . Therefore, $ {T_1 ^2 \over T_2^2 } = .........( k =1 m/s^2 and g = 10 m/s^2 ) $

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Explanation

$ here y = kt^2 $ $ \therefore { dy \over dt } = 2 kt \Rightarrow { d^2 y \over dt^2 } = 2k = 2 ms^{-2} $ $ \therefore the point of support in \;moving\; upwards with\; an acceleration\; of 2 m/s^2 $ $ \therefore effective acceleration g' = g + a = 12 m/s^2 $ $ now \;T_1 = 2 \pi \sqrt { l \over g } and\; T_2 = 2 \pi \sqrt { l \over g } $

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A hollow sphere is filled with water. There is a hole at the bottom of this sphere. This sphere is suspended with a string from a rigid support and given an oscillation. During oscillation, the hole is opened up and the periodic time of this oscillating system is measured. The periodic time of the system………….

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Explanation

$ T = 2 \pi \sqrt { l \over g } $ as water leaks, the center of gravity moves down and hence “ l ” increases. $ \therefore $ T increases initially When all the water has leaked, the center of gravity moves up and hence “ l ” decreases and hence T decreases Finally the centre of gravity steady at the center of sphde and so T will remain constant.

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The periodic time of a S.H.O. oscillating about a fixed point is 2 s. After what time will the kinetic energy of the oscillator become $ 25 \% $ of its total energy?

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Explanation

Kinetic energy = 25 % E $ \therefore K = { 1 \over 4 } E $

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A body having mass 5g is executing S.H.M. with an amplitude of 0.3 m. If the periodic time of the system is $ { \pi \over 10 } s $ , then the maximum force acting on body is ……….

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Explanation

$ F_{max} = ma_{max} =mA \omega ^2 = mA { 4 \pi^2 \over T^2 } = 0.6 N $

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A particle is executing S.H.M. between x= - A and x = +A. If the time taken by the particle to travel from x = 0 to A/2 is $ T_1 $ T1 and that taken to travel from x = A/2 to x = A is $ T _2 $ , then

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Explanation

$ \nu = w \sqrt { A^2 -x^2 } $ the velocity for moving form x=o to x = A /2 will ge more themfor x =A /2 to x = A $ \therefore T_1 \lt T_2 $

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For a particle executing S.H.M., when the potential energy of the oscillator becomes 1/8 the maximum potential energy, the displacement of the oscillator in terms of amplitude Awill be

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Explanation

$ U = {1 \over 8 } U_{max} $ $ \therefore {1 \over 2} ky^2 = { 1 \over 8 } \left( {1 \over 2 } kA^2 \right) \Rightarrow y^2 = { A^2 \over 8 } $

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The average values of potential energy and kinetic energy over a cycle for a S.H.O. will be ……………….. respectively.

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Explanation

In the expression for both Kinetic and potential energy, We have the square of the halmonic functions (sine or cisine). The average of which over a cycle is 12 $ \therefore \lt u \gt = { E \over 2 } = \lt K \gt = { 1 \over 4} m \omega^2 A^2 $

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The ratio of force constants of two springs is 1:5. The equal mass suspended at the free ends of both springs are performing S.H.M. If the maximum acceleration for both springs are equal, the ratio of amplitudes for both springs is ………

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Explanation

$ Angular frequency \omega = \sqrt { k \over m} $ $ Since 'm' is constant , \omega \alpha \sqrt k ^1 $ $Now , a _{max} = A \omega^2 \Rightarrow \omega = \sqrt { a_{max} \over A } $ $ \therefore {a_{max} \over A} = k \Rightarrow { a_{max} \over K } = A $ $ \therefore A \alpha {1 \over k } $

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When a mass M is suspended from the free end of a spring, its periodic time is found to be T. Now, if the spring is divided into two equal parts and the same mass M is suspended and oscillated, the periodic time of oscillation is found to be T’. Then

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Explanation

$ For a spring , T = 2 \pi \sqrt { m \over k } \Rightarrow T \alpha { 1 \over \sqrt k } $ ( m is constant )

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The periodic time of two oscillators are T and 5T/ 4 respectively. Both oscillators starts their oscillation simultaneously from the mid point oftheir path of motion. When the oscillator having periodic timeT completes one oscillation, the phase difference between the two oscillators will be ………

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Explanation

$ Phase of 1st oscillator \theta_1 = \omega _1 t + \phi = { 2 \phi \over T_1 } t + \phi $ $ For 2nd oscillator , \theta_2 = \omega_2 t + \phi = {2 \phi \over T_2 } t + \phi $ $ Phase diff \theta_1 - \theta_2 $

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