A particle is executing SHM with time period T. The time taken by it to travel from mean position to times its amplitude is equal to
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A block of mass 2 kg is hanging with a massless spring and the spring is stretched by 40 cm. If the block is pulled down and released, then the period of oscillation is: (here, g = 10 )
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A spring is attached vertically to the ceiling of a lift and the lower end of spring is connected with a block of mass \(2~\text{kg}\). If the lift starts accelerating upwards with an acceleration \(2~\text{m/s}^2,\) then find the amplitude of SHM, while the spring constant is \(100~\text{N/m}\):
1. \(8~\text{cm}\)
2. \(1~\text{cm}\)
3. \(2~\text{cm}\)
4. \(4~\text{cm}\)
The motion of the particle is started at t = 0 and the equation of motion is given by , where x is in cm and t is in seconds. When will the particle come to rest for the first time?
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What will be the frequency of oscillation of a simple pendulum, if the length of the pendulum is equal to the radius of earth?
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The graph of potential energy \((U)\) versus displacement \((x)\) is shown. Which of the following describes the oscillation about the mean position, \(x = 0\text{?}\)
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The time period of a body under S.H.M. is and when restoring forces and respectively act on it. What will be the time period of S.H.M. when both the forces act simultaneously on it?
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4. \(\frac{\mathrm{T_1 T_2}}{\sqrt{\mathrm{T^2_1+T^2_2}}}\)
A particle is subjected to two mutually perpendicular SHM such that x = 2sint and
. The path of the particle will be
1. An ellipse
2. A straight line
3. A parabola
4. A circle
When a periodic force \(\vec{F_1}\) acts on a particle, the particle oscillates according to the equation \(x=A\sin\omega t\). Under the effect of another periodic force \(\vec{F_2}\), the particle oscillates according to the equation \(y=B\sin(\omega t+\frac{\pi}{2})\). The amplitude of oscillation when the force (\(\vec{F_1}+\vec{F_2}\)) acts are:
1. | \(A+B\) | 2. | \(\sqrt{A^2+B^2}\) |
3. | \(\large\frac{\sqrt{A^2+B^2}}{2}\) | 4. | \(\sqrt{AB}\) |