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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 ……….
$ F_{max} = ma_{max} =mA \omega ^2 = mA { 4 \pi^2 \over T^2 } = 0.6 N $
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
$ \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 $
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
$ 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 } $
The average values of potential energy and kinetic energy over a cycle for a S.H.O. will be ……………….. respectively.
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 $
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 ………
$ 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 } $
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
$ For a spring , T = 2 \pi \sqrt { m \over k } \Rightarrow T \alpha { 1 \over \sqrt k } $ ( m is constant )
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 ………
$ 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 $
A rectangular block having mass mand cross sectional area A is floating in a liquid having density r. If this block in its equilibrium position is given a small vertical displacement, its starts oscillating with periodic time T. Then in this case…..
$ Restoring force F = - Ay \rho g = -( A \rho g ) y = -ky $ $ \therefore k = A \rho g \Rightarrow T = 2 \pi \sqrt { m \over k } \Rightarrow T \alpha { 1 \over sqrt A^1 } $
Which of the equation given below represents a S.H.M.?
In SHM, acceleration and displacement are opposite in direction Also $ a \alpha y $.
The displacement for a particle performing S.H.M. is given by $ x = A Cos( ùt + \hat O) $ . If the initial position of the particle is 1 cm and its initial velocity is $p cms^{- 1} $ , thenwhat will be its initial phase? The angular frequency of the particle is $p s^{-1}.$
$ Here t =0 , x =1 cm and \nu = \pi cm s^{-1} , w = \pi s^{-1} $ $ Now , x = A cos ( \omega t + \phi ) ....(1) $ $ Velocity \nu = { dx \over dt } = -A sin \omega ( \omega t + \phi ) .....(2) $ Solved the equation (1) and (2)
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.