Practice free Oscillations (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
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
$ 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 $
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 ………….
$ \omega = \sqrt { k \over m_1 + m_2 } or\, removing \,m , angular\, frequency \,\omega' = { k \over m_2 } $
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 }
$ 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 $
The speed of a particle executing motion changes with time according to the equation $y = aSinùt + bCosùt,$ then ……..
$ 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 } $
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?
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} $
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 ……….
$ 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 $
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?
$ 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 $
When a mass m is suspended from the free end of a massless spring having force constant k, its oscillates with frequency f. Now if the spring is divided into two equal parts and a mass 2m is suspended from the end of anyone of them, it will oscillate with a frequency equal to ………….
$ f = { 1 \over 2 \pi } \sqrt { k \over m } and f' = { 1 \over 2 \pi } \sqrt { 2k \over 2m } $ $ { k' = 2 k } $ $ \therefore f' = f $
A body of mass 1 kg suspended from the free end of a spring having force constant $400 Nm^{-1}$ is executing S.H.M. When the total energy of the system is 2 joule, the maximum acceleration is
………$ms^{ – 2}$ .
Energy stoved =Work done $ \therefore E = { 1 \over 2 } k A^2 $ Now maximum acceleration $ a_{max} = \omega^2 A $
A spring is attached to the center of a frictionless horizontal turn table and at the other end a body of mass 2 kg is attached. The length of the spring is 35 cm. Now when the turn table is rotated with an angular speed of $10 rad s^{– 1}$ , the length of the spring becomes 40 cm then the force constant of the spring is..... N/m.
Radius of the rotational motion r =0.4 m When the turn table rotates, the restoring force developed in the spring = centrifugal force $ \therefore F_{restore} = m \omega^2 r = 2 ( 10 ) ^2 \times 0.4 = 80 N $ Now increase in length of spring = 40-35 = 5 cm $ \therefore Force constant k - { F \over x } = { 80 \over 0.05 } = 1.6 \times 10^3 N/m $
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