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The equation for displacement of a particle at time t is given by the equation y = 3Cos2t + 4Sin2t The periodic time of oscillation is ………
$ T = { 2 \pi \over \omega } = {2 \pi \over 2 } = \pi s $
The equation for displacement of a particle at time t is given by the equation y = 3Cos2t + 4Sin2t. . The amplitude of oscillation is ……cm
$ amplitude A = \sqrt { 3^2 + 4^2 } = 5 cm $
The equation for displacement of a particle at time t is given by the equation y = 3Cos2t + 4Sin2t. . The maximum acceleration of the particle is…………cm / s2.
Maximum Accelaration of $ A particle = A \omega^2 = 5 (2)^2 = 20 cms^{-2} $
The equation for displacement of a particle at time t is given by the equation y = 3Cos2t + 4Sin2t. . If the mass of the particle is 5 gm, then the total energy of the particle is ……erg.
$ Mechanical energy = {1 \over 2 } m \omega ^2 A^2 = 250 erg $
The equation for displacement of a particle at time t is given by the equation y = 3Cos2t + 4Sin2t. . The frequency of the particle is ………s- 1 .
$ Frequenct of the particle f = {1 \over T } = { 1 \over \pi } s^{-1} $
Equation for a harmonic progressive wave is given by y = Asin ( 15pt + 10px + p/3) where x is in meter and t is in seconds. This wave is ……….
$ on comparing y = A sin ( 15 \pi t + 10 \pi x + { \pi \over 3 }) $ $ with y = A sin ( \omega t + kx + \theta ) $
If the velocity of sound wave in humid air is $v_m$ and that in dry air is $v_d$, then……
At constant pressure density of water vapour is less than dry air. $ \therefore $ with increase in humidityaccording to the equation $ \nu = \sqrt { \gamma p \over \rho } $ the velocity of sound increases.
The ratio of frequencies of two waves travelling through the same mediumis 2:5. The ratio of their wavelengths will be ………
$ f \alpha \lambda^{-1} $ $ \therefore { f_1 \over f_2} = { \lambda_2 \over \lambda_1 } $
If the maximum frequency of a sound wave at room temperature is 20,000 hz then its minimum wavelength will be approximately…….$ ( \nu = 340 ms^{-1} $
From the equation $ v = f \lambda , \lambda_{min} = { \nu \over f_{max} } = 17 mm $ which is nearer to 20 mm
If the equation of a wave in a string having linear mass $ 0.04 kg m^{-1} $ is given by $ y = 0.02 sin \left[ 2 \pi \left( {t \over 0.04 } - { x \over 0.50 } \right) \right] $ , then the tension in the string is……….. N. ( All values are in mks )
On comparing with the wave equation $ y = A sin 2 \pi \left( { t \over T } - { x \over \lambda} \right) we get , T = 0.04 s , \lambda = 0.5 m \Rightarrow \nu = { 25 \over 2 } ms^{-1} $ $ \therefore T = \nu^2 \pi = 6.25 N $
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.