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, \(x = 0\text{?}\)
1. | 2. | ||
3. | 4. |
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. | \(\sqrt{AB}\) |
A particle is executing SHM with amplitude A and angular frequency . The time taken by the particle to move from x = 0 to x = A/2 is
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A spring-block system oscillates with a time period \(T\) on the earth's surface. When the system is brought into a deep mine, the time period of oscillation becomes \(T'.\) Then one can conclude that:
1. \(T'>T\)
2. \(T'<T\)
3. \(T'=T\)
4. \(T'=2T\)
A particle executes simple harmonic oscillations under the effect of small damping. If the amplitude of oscillation becomes half of the initial value of 16 mm in five minutes, then what will be the amplitude after fifteen minutes?
1. 8 mm
2. 4 mm
3. 2 mm
4. 1 mm