Oscillations MCQs for NEET — Physics Questions with Answers

Here T=6 s Amplitude OB = OC = 1/2 BC = 10cm $ \therefore OD =5 cm$ $ Now displacement x = A sin ( wt + \phi) ...(1) $ $ where A = 10 gm , \omega = { 2 \pi \over T } = { \pi \over 3 } rad $ Now if at t = 0 , oscillator is at C i.e at t =0 , x = A $\therefore A= Asin(\omega \times 0 + \phi) (OR) A=Asin \phi$ $ \Rightarrow sin \phi = 1 $ $ \Rightarrow \phi ={\pi \over 2} $ putting this in eqn (1) $ x = A sin ( \omega t + { \pi \over 2 } ) = A cos \omega t = 10 cos \omega t $ $ \therefore for x = 5 cm $ $ 5 = 10 cos \omega t \Rightarrow cos \omega t = { 1 \over 2 } $ $ \therefore \omega t = { \pi \over 3 } $ $ \therefore t = 1 S $

Force responsible for oscillation in $ F = mg sin \theta = mg \theta $ $ \{ \theta is small \} $ $ = mg .{ x \over R} $ Comparing this with $ F = - kx $ $ k = { mg \over R } $

Here 2 acceleration vectors g. and a are acting along mutually prependicular direction . $ \therefore effective acceleratioin l^n g_{eff} = \sqrt { g^2 + a^2 } $ $ \therefore T = 2 \pi \sqrt { l \over g _{eff} } $

$ g^2 _{off} = a_x^2 + (g -ay ) ^2 here a_x g sin \theta cos \theta , ay = g sin ^2 \theta $ $ = a_x^2 + g^2 + a_y^2 - 2 ga_y $ $ = a^2 sin ^ 2 \theta cos ^2 \theta + g^2 + g^2 sin^2 \theta - 2 g^2 sin ^2 \theta $ $ = g^2 ( 1 -sin^2 \theta ) $ $= g ^2 cos ^2 \theta $ $ \therefore g_{eff} = g cos \theta $

$ E_1 = {1 \over 2 } m \omega^2 x^2 \Rightarrow \sqrt E_1 = x \sqrt { {1 \over 2 } m \omega^2 } ...(1) $ $ E_2 = {1 \over 2 } m \omega^2 y^2 \Rightarrow \sqrt E_2 = x \sqrt { {1 \over 2 } m \omega^2 } ...(2) $ $ E = {1 \over 2 } m \omega^2 (x + y )^2 \Rightarrow \sqrt E = (x + y ) \sqrt { {1 \over 2 } m \omega^2 } ...(3) $ From (1) ,(2) ,(3) , $ \sqrt E = \sqrt E_1 + \sqrt E_2 $ $or E = E_1 + E_2 + 2 \sqrt { E_1 E_2 } $

$ \omega_1^2 = { k \over m } = { kx \over mx } = { F_1 \over mx} ...(1) $ $ similarly , \omega_2^2 = { F_2 \over mx} ...(2) $ $ if F_1 and F_2 acts simultaneously ,then angular frequency $ $ w_2 ={ F_1 + F_2 \over mx } .....(3) $ $ From (1) , (2) and (3) ; \omega^2 = \omega_1^2 + \omega_2^2 $ $ now ,use eqn \omega = { 2 \pi \over T} $

Initial periodic time $T_1 = 2 \pi \sqrt { l \over g} …..(1) $ When pendulum moves along vertical direction, effective acceleration $ g_{eff} = g+a $ where ‘a’ inaccleration of pendulum. $now , a = { d \nu \over dt} = { d (kt) \over dt } = k = 2.1 ms^{-2} $ $ \therefore New periodic time T_2 = 2 \pi \sqrt { l \over g_{eff} } .....(2) $ $ \therefore { T_2 \over T_1 } = \sqrt { g \over g_{eff} }^1$

Block will not slide if $ \mu mg \geq ma \Rightarrow \mu g \geq a $ To prevent the block from sliding the maximum acceleration of table must be $a_max = \mu g$ Now maximum accleration $ a_{max} = \omega^2 A $ $ \mu^2 A_{max} = \mu g $ $ \therefore A_{max} = { \mu g \over \omega^2 }= { \mu g T^2 \over 4 \pi^2 } $

At the upper most end, $ when mg = R + m \omega^2A$ coin will loose contact. Taking R=0 $ m \omega^2 A = mg $ $ A = { g / \omega ^ 2} $

Force required to increase the length by x in F = kx….. (1) After spring is divided into 2 equal parts, F = k ' x ' where x ' = x/2 = k' x/2 …..(2) From (1) and (2) ; k' = 2 k