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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 } $
A hollow sphere is filled with water. There is a hole at the bottom of this sphere. This sphere is suspended with a string from a rigid support and given an oscillation. During oscillation, the hole is opened up and the periodic time of this oscillating system is measured. The periodic time of the system………….
$ T = 2 \pi \sqrt { l \over g } $ as water leaks, the center of gravity moves down and hence “ l †increases. $ \therefore $ T increases initially When all the water has leaked, the center of gravity moves up and hence “ l †decreases and hence T decreases Finally the centre of gravity steady at the center of sphde and so T will remain constant.
The periodic time of a S.H.O. oscillating about a fixed point is 2 s. After what time will the kinetic energy of the oscillator become $ 25 \% $ of its total energy?
Kinetic energy = 25 % E $ \therefore K = { 1 \over 4 } E $
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