Acceleration of point x = 2m of a transverse wave is 5 j ^ m/s^2 and the slope of the waveform is given by dy/dx = 5sin (pi*x)/12 . Then the speed of the point at that instant is given in m/s as

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Abhishek Kumar · Answer

V p = -V y dy/dx V p = Velocity of the particle V y = Velocity of the wave Solve.

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Acceleration of point x = 2m of a transverse wave is 5 j ^ m/s^2 and the slope of the waveform is given by dy/dx = 5sin (pi*x)/12 . Then the speed of the point at that instant is given in m/s as

Acceleration of point x = 2m of a transverse wave is 5 j^{^} m/s^2 and the slope of the waveform is given by dy/dx = 5sin (pi*x)/12 . Then the speed of the point at that instant is given in m/s as

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Acceleration of point x = 2m of a transverse wave is 5 j ^ m/s^2 and the slope of the waveform is given by dy/dx = 5sin (pi*x)/12 . Then the speed of the point at that instant is given in m/s as

Acceleration of point x = 2m of a transverse wave is 5 j^ m/s^2 and the slope of the waveform is given by dy/dx = 5sin (pi*x)/12 . Then the speed of the point at that instant is given in m/s as

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Acceleration of point x = 2m of a transverse wave is 5 j^{^} m/s^2 and the slope of the waveform is given by dy/dx = 5sin (pi*x)/12 . Then the speed of the point at that instant is given in m/s as