NEET Physics Questions: Oscillations

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The distance travelled by a particle performing S.H.M. during time interval equal to its periodic time is ……..




In Simple Harmonic Motion (SHM), the distance traveled by a particle in one complete cycle (periodic time \( T \)) is four times the amplitude \( A \). This is because the particle travels from one extreme to the other extreme, back to the initial extreme, and then to the starting point, covering a total distance of \( 4A \).
A person standing in a stationary lift measures the periodic time of a simple pendulum inside the lift to be equal to T. Now, if the lift moves along the vertically upward direction withan acceleration of g/3 ,then the periodic time of the lift will now be ………




$ T = 2 \pi \sqrt { l \over g } $ When lift moves up with accleration g/3 the effective graritatianl acclenations in $ g^1 = g + { g \over 3} = {4g \obver 3 } $ $ \therefore new peliodic time T ' = 2 \pi \sqrt { l \over g } $
If the equation for displacement of two particles executing S.H.M. is given by $y_1 = 2Sin(10t+è)$ and $y_2 = 3Cos10t$ respectively, then the phase difference between the velocity of two particles will be ………..




$ v_1 = { dy_1 \over dt } = 2 \times 10 cos ( 10t + \theta ) $ $ v_2 = -3 \times 10 sin t = 30 cos ( 10 + { \pi \over 2 } ) $ $ \therefore Phase difference = (10 t + \theta) - ( 10 + { \pi \over 2 } ) = \theta - { \pi \over 2 } $
If the maximum velocity of two springs ( both has same mass ) executing S.H.M. and having force constants $ k_1 and k_2 $ respectively are same, then the ratio of their amplitudes will be




For two springs with the same mass executing SHM, the maximum velocity \( v_{max} \) is given by \( v_{max} = A \omega \), where \( A \) is the amplitude and \( \omega \) is the angular frequency. Since \( \omega = \sqrt{\frac{k}{m}} \), and the maximum velocities are the same, we have \( A_1 \sqrt{\frac{k_1}{m}} = A_2 \sqrt{\frac{k_2}{m}} \). Simplifying, we get \( \frac{A_1}{A_2} = \sqrt{\frac{k_2}{k_1}} \). Therefore, the ratio of their amplitudes is \( \sqrt{\frac{k_2}{k_1}} \).
The bob of a simple pendulum having length ‘ l’ is displaced from its equilibrium position by an angle of è and released. If the velocity of the bob, while passing through its equilibrium position is v, then v =




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.
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If 1/4 of a spring having length l is cutoff, then what will be the spring constant of remaining part?




$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 $
The amplitude for a S.H.M. given by the equation x = 3Sin3pt + 4Cos3pt is ………m.




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 $
When an elastic spring is given a displacement of 10mm, it gains an potential energy equal to U. If this spring is given an additional displacement of 10 mm, then its potential energy will be




$ u \alpha y^2 $ $ \therefore { u_2 \over u_1} = \left( { y_2 \over y_1} \right)^2 \Rightarrow u_2 = 4u $
The increase in periodic time of a simple pendulum executing S.H.M. is............ when its length is increased by $ 21 \% $ .




$ 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 \% $
A particle executing S.H.M. has an amplitude A and periodic time T. The minimum time required by the particle to get displaced by $ A / \sqrt 2 $ from its equilibrium position is ...................s.




$ 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 } $
If a body having mass M is suspended from the free ends of two springs A and B, their periodic time are found to be $T_1$ and $T_2$ respectively. If both these springs are now connected in series and if the same mass is suspended from the free end, then the periodic time is found to be T. Therefore …………..




$ 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 $
The displacement of a S.H.O. is given by the equation $ x = A cos ( ut + { \pi \over 8 } ) $ . At what time will it attain Maximum velocity?




$ 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 }$
At what position will the potential energy of a S.H.O. become equal to one third its kinetic energy?




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 } $
For particles A and B executing S.H.M., the equation for displacement is given by $y_1 = 0.1Sin(100t+p/3) $ and $y_2 = 0.1Cospt $ respectively. The phase difference between velocity of particle A with respect to that of B is …………




$ \nu_1 = { dy_1 \over dt } $ $ and \nu_2 = { dy_2 \over dt } $
The periodic time of a simple pendulum is $T_1$. Now if the point of suspension of this pendulum starts moving along the vertical direction according to the equation $y = kt^2$, the periodic time of the pendulum becomes $T_2$ . Therefore, $ {T_1 ^2 \over T_2^2 } = .........( k =1 m/s^2 and g = 10 m/s^2 ) $




$ here y = kt^2 $ $ \therefore { dy \over dt } = 2 kt \Rightarrow { d^2 y \over dt^2 } = 2k = 2 ms^{-2} $ $ \therefore the point of support in \;moving\; upwards with\; an acceleration\; of 2 m/s^2 $ $ \therefore effective acceleration g' = g + a = 12 m/s^2 $ $ now \;T_1 = 2 \pi \sqrt { l \over g } and\; T_2 = 2 \pi \sqrt { l \over g } $