NEET Practice Questions (MCQs) with Answers & Solutions

The velocity of the bob of a simple pendulum when it passes through its equilibrium position can be derived using the principle of conservation of mechanical energy. At the maximum displacement (angle θ), all the energy is potential, and at the equilibrium position, all the energy is kinetic. Therefore, we equate the potential energy at the maximum displacement to the kinetic energy at the equilibrium position. The formula for the velocity (v) is given by: $$ v = \\sqrt{2gl(1 - \\cos \\theta)} $$ where g is the acceleration due to gravity, l is the length of the pendulum, and θ is the angle of displacement.

$K_1 l _ 1 = K_2 l_2 = kl $ $ \therefore K_1 \left( l \over 4 \right) = K_2 \left( { 3 \over 4 } l \right) = kl $ $ force constant of spring having length {3 \over 4 } l \;in $ $ k_2 = { 4 \over 3 } k $

Amplitude of SHM given by $x = a sin \omega t+b cos \omega t $in $ A = \sqrt { a^2 + b^2 } = ( 3^2 + 4 ^2 ) ^ { 1/2 } = 5 m $

$ u \alpha y^2 $ $ \therefore { u_2 \over u_1} = \left( { y_2 \over y_1} \right)^2 \Rightarrow u_2 = 4u $

$ T \alpha \sqrt 1 ^1 $ $ because \,2 \pi and\, g are \,constants $ $ \therefore { T_2 \over T_1} = \sqrt { l_2 \over l_1 } = \sqrt {1.2 l_1 \over l_1 } =1.1 $ $ \therefore \% increase = { T_2 -T_1 \over T_1 } \times 100 = 10 \% $

$ y = A sin ( \omega t + \phi ) $ $ { A \over \sqrt 2 } = A sin \omega t \{ \phi = 0 \} \Rightarrow { 1 \over \sqrt 2 } = sin \omega t = { \pi \over 4 } $ $ \therefore { 2 \pi \over T } .t = { \pi \over 4 } $ $ \therefore t = { T \over 8 } $

$ T_1 = 2 \pi \sqrt { m^1 \over k_1 } \Rightarrow k_1 = { 4 \pi^2 M \over T_1^2 } and k_2 = { 4 \pi ^2 M \over T_2^2 } $ For series connection ; $ T = 2 \pi \sqrt { M \over k} where k = { k_1 k_2 \over k_1 + k_2 } $ $ \therefore T = 2 \pi \sqrt { {M^1 \over 4 \pi^2 M} ( T_1^2 + T_2 ^2 ) } $ $ \therefore T = \sqrt { T_1^2 + T_2 ^2 } \Rightarrow T^2 = T_1 ^2 + T_2 ^2 $

$ x = A cos ( \omega t + { \pi \over 8 } ) \Rightarrow \nu = { dx \over dt } = - A \omega sin ( \omega t + { \pi \over 2 } ) $ if $ sin ( \omega t + { \pi \over 8 } ) = 1$ , then velocity will be maximum $ \Rightarrow \omega t + { \pi \over 8 } = { \pi \over 8 } \Rightarrow \omega t = { 3 \pi \over 8 } \Rightarrow t = { 3 \pi \over 8 \omega }$

3 u = k $ \therefore 3 \times { 1 \over 2 } ky^2 = { 1 \over 2 } k ( A^2 - y^2 ) \Rightarrow y = \pm { A \over 2 } $

$ \nu_1 = { dy_1 \over dt } $ $ and \nu_2 = { dy_2 \over dt } $