A particle starts with $S.H.M.$ from the mean position as shown in the figure. Its amplitude is $A$ and its time period is $T$. At one time, its speed is half that of the maximum speed. What is this displacement?
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In figure $(A),$ mass ' $2 m$ ' is fixed on mass ' $m$ ' which is attached to two springs of spring constant $k$. In figure $(B),$ mass ' $m$ ' is attached to two spring of spring constant ' $k$ ' and ' $2 k$ '. If mass ' $m$ ' in $(A)$ and $(B)$ are displaced by distance ' $x$ ' horizontally and then released, then time period $T_{1}$ and $T_{2}$ corresponding to $(A)$ and $(B)$ respectively follow the relation.
Two particles oscillating in $SHM$ along two very close parallel path such that they have same mean position. The equation of $SHM$ of two particles are $x_1 = A\, sin\,\omega t$ and $x_2 = A\,sin(\omega t + \phi )$ respectively. If maximum distance between them is $\frac{6A}{5}$ then $\phi $ equal to ..... $^o$
The potential energy of a particle of mass $4\,kg$ in motion along the $x$-axis is given by $U =4(1-\cos 4 x )\,J$. The time period of the particle for small oscillation $(\sin \theta \simeq \theta)$ is $\left(\frac{\pi}{ K }\right)\,s$. The value of $K$ is .......
Two oscillating systems; a simple pendulum and a vertical spring-mass-system have same time period of motion on the surface of the Earth. If both are taken to the moon, then-