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Physics MCQs for NEET — Practice Questions with Answers

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Two simple pendulums having lengths 144 cm and 121 cm starts executing oscillations. At some time, both bobs of the pendulum are at the equilibrium positions and in same phase. After how many oscillations of the shorter pendulum will both the bob’s pass through the equilibrium position and will have same phase?

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Explanation

$ T_1 = 2 \pi \sqrt { 144 \over g } and \,T_2 = 2 \pi \sqrt { 121 \over g } $ $ \therefore T_1 \gt T_2 $ When the shorter pendulum completes n oscillations, the longer one completes (n-1) oscillations (when in same phase). $ \therefore nT_2 = (n -1 ) T_1 $

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The maximum velocity and maximum acceleration of a particle executing S.H.M. are 1 m/s and $3.14 m/s^2$ respectively. The frequency of oscillation for this particle is ……..

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Explanation

$ \therefore \omega = { \omega^2 r \over r \omega } = 3.14 $ $ \therefore 2 \pi f = 3.14 \Rightarrow f = { 3.14 \over 2 \pi } = 0.5 s^{-1} $

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A particle having mass 1 kg is executing S.H.M. with an amplitude of 0.01 mand a frequency of 60 hz. The maximum force acting on this particle is………… N

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Explanation

$ Maximum force = m \omega ^2 A = m 4 \pi^2 f^2 A $

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A simple pendulum having length l is given a small angular displacement at time t = 0 and released. After time t, the linear displacement of the bob of the pendulum is given by

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Explanation

$ Periodic time T = 2 \pi { l \over g } = \omega = {2 \omega \over T } \Rightarrow \omega = { g \over l } $ $ Linear \,displacement x = a cos \omega t $

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Two masses $m_1$ and $m_2$ are attached to the two ends of a massless spring having force constant k. When the system is in equilibrium, if the mass $m_1$ is detached, then the angular frequency of mass $m_2$ will be ………….

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Explanation

$ \omega = \sqrt { k \over m_1 + m_2 } or\, removing \,m , angular\, frequency \,\omega' = { k \over m_2 } $

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When the displacement of a S.H.O. is equal to A/2, what fraction of total energy will be equal to kinetic energy? { Ais amplitude }

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Explanation

$ Kinetic energy K = { 1 \over 2 } m \omega^2 ( A^2 - y^2 ) $ $ Now , Total energy E = { 1 \over 2 } m \omega^2 A^2 $

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The speed of a particle executing motion changes with time according to the equation $y = aSinùt + bCosùt,$ then ……..

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Explanation

$ y = a sin \omega t + b cos \omega t $ $ Taking a = A cos \theta and b = A sin \theta $ $ y = A cos \theta sin \omega t + A sin \theta cos \omega t $ $ = A sin ( \omega t + \theta ) $ $ Now , a^2 + b^2 = A^2 $ $ \therefore A = \sqrt { a^2 + b^2 } $

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A body is placed on a horizontal plank executing S.H.M. along vertical direction. Its amplitude of oscillation is $3.92 \times 10^{– 3}m$. What should be the minimum periodic time so that the body does not loose contact with the plank?

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Explanation

The body will not loose contact with the surface, $ if mg = m \omega ^2 r = { m4 \pi ^2 \over T^2 } . r $ (where r is amplitude ) $ \therefore T = 2 \pi \sqrt { r \over g} $

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If the kinetic energy of a particle executing S.H.M. is given by K = K Cos2ùt, then the displacement of the particle is given by ……….

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Explanation

$ maximum kinetic energy K_o = { 1 \over 2 } m \omega^2 A^2 $ $ \therefore A = \left( { 2 K_o \over m \omega^2 } \right) ^{1 /2 } $ $ \therefore Equation for displacement is ; y = A sin \omega t = \left( { 2 K_o \over m \omega^2 } \right) ^{1 /2} sin \omega t $

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The equation for displacement of two identical particles performing S.H.M. is given by $x_1 =4Sin(20t+p/6)$ and $x_2 =10Sinùt$. For what value of ù will both particles have same energy?

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Explanation

$ E = { 1 \over 2 } m \omega^2 A^2 \Rightarrow E \alpha \omega^2 A^2 $ $ \therefore E \alpha ( A \omega ) ^2 \Rightarrow ( \omega_1 A_1 ) ^2 = ( \omega_2 A_2 ) ^2 $

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