If a watch with a wound spring is taken on to the moon, it
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(d)The time period of oscillation of a spring does not depend on gravity.
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The potential energy of a simple harmonic oscillator of mass $2\, kg$ in its mean position is $5\, J.$ If its total energy is $9\,J$ and its amplitude is $0.01\, m,$ its time period would be
If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is $\frac{x}{2}$ times its original time period. Then the value of $x$ is:
Four harmonic waves of equal frequencies and equal intensities $I_0$ have phase angles $0, \pi / 3,2 \pi / 3$ and $\pi$. When they are superposed, the intensity of the resulting wave is $nI _0$. The value of $n$ is
Figure shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and the initial position are indicated on the figure. The simple harmonic motion of the $x-$ projection of the radius vector of the rotating particle $P$ is
$Y = A \sin \left(\omega t +\phi_{0}\right)$ is the time-displacement equation of a SHM. At $t=0$ the displacement of the particle is $Y =\frac{ A }{2}$ and it is moving along negative $x$ -direction. Then the initial phase angle $\phi_{0}$ will be ...... .
Time period of a simple pendulum is $T$. The angular displacement for amplitude is $\beta$. How much time the bob of pendulum will take to move from equilibrium position $O$ to $A$, making an angle $\alpha$ at the support
A simple harmonic motion is represented by $y\, = 5\,(\sin \,3\pi t\, + \,\sqrt 3 \,\cos \,3\pi t)\,cm$ The amplitude and time period of the motion are
An object of mass $0.5\, {kg}$ is executing simple harmonic motion. Its amplitude is $5\, {cm}$ and time period (T) is $0.2\, {s} .$ What will be the potential energy of the object at an instant $t=\frac{T}{4}$ s starting from mean position. Assume that the initial phase of the oscillation is zero. (In ${J}$)