A simple harmonic motion is represented by $F(t) = 10\sin \,(20\,t + 0.5)$. The amplitude of the $S.H.M.$ is $a$ $=$ ....
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Two simple pendulums of length $0.5\, m$ and $2.0\, m$ respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed .... oscillations.
A point performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x=Asin$$\left( {\omega t + \frac{\pi }{6}} \right)$. After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity?
The composition of two simple harmonic motions of equal periods at right angle to each other and with a phase difference of $\pi $ results in the displacement of the particle along
A particle at the end of a spring executes simple harmonic motion with a period ${t_1}$, while the corresponding period for another spring is ${t_2}$. If the period of oscillation with the two springs in series is $T$, then
A particle performs simple harmonic motion with a period of $2$ second. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $\frac{1}{ a } s .$ The value of $'a'$ to the nearest integer is.........
The displacement time graph of a particle executing $S.H.M.$ is given in figure: (sketch is schematic and not to scale) Which of the following statements is are true for this motion?
$(A)$ The force is zero $t=\frac{3 T}{4}$
$(B)$ The acceleration is maximum at $t=T$
$(C)$ The speed is maximum at $t =\frac{ T }{4}$
$(D)$ The $P.E.$ is equal to $K.E.$ of the oscillation at $t=\frac{T}{2}$
A particle is executing $SHM$ with amplitude $A,$ time period $T,$ maximum acceleration $a_o$ and maximum velocity $v_0.$ Its starts from mean position at $t=0$ and at time $t,$ it has the displacement $A/2,$ acceleration $a$ and velocity $v$ then