Questions in Oscillations

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A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position the velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is
Two particles P and Q start from origin and execute Simple Harmonic Motion along X-axis with same amplitude but with periods 3 seconds and 6 seconds respectively. The ratio of the velocities of P and Q when they meet is
A particle is performing simple harmonic motion with amplitude $A$ and angular velocity $\omega$. The ratio of maximum velocity to maximum acceleration is
The angular velocities of three bodies in simple harmonic motion are ${{\omega }_{1}},\,{{\omega }_{2}},\,{{\omega }_{3}}$ with their respective amplitudes as ${{A}_{1}},\,{{A}_{2}},\,{{A}_{3}}$. If all the three bodies have same mass and velocity, then
The velocity of a particle performing simple harmonic motion, when it passes through its mean position is
The velocity of a particle in simple harmonic motion at displacement $y$ from mean position is
A particle is executing the motion $x=A\cos (\omega \,t-\theta )$. The maximum velocity of the particle is
A particle executing simple harmonic motion with amplitude of $0.1 m$. At a certain instant when its displacement is $0.02 m$, its acceleration is $0.5 ms^{-2}$. The maximum velocity of the particle is (in m/s)
The amplitude of a particle executing SHM is $4 cm$. At the mean position the speed of the particle is $16 cm/sec$. The distance of the particle from the mean position at which the speed of the particle becomes $8\sqrt{3}cm/s,$will be
The maximum velocity of a simple harmonic motion represented by $y=3\sin \,\left( 100\,t+\frac{\pi }{6} \right)$is given by

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