Let $T_1$ and $T_2$ be the time periods of two springs $A$ and $B$ when a mass $m$ is suspended from them separately. Now both the springs are connected in parallel and same mass $m$ is suspended with them. Now let $T$ be the time period in this position. Then
$T = T_1+ T_2$
$T = \frac{T_1T_2}{T_1+ T_2}$
$T^2 = T_1^2 + T_2^2$
$\frac{1}{T^2} =\frac{1}{T_1^2}+\frac{1}{T_2^2}$
A mass $M$, attached to a horizontal spring, executes S.H.M. with amplitude $A_1$. When the mass $M$ passes through its mean position then a smaller mass $m$ is placed over it and both of them move together with amplitude $A_2$. The ratio of $\frac{{{A_1}}}{{{A_2}}}$ is
A spring executes $SHM$ with mass of $10\,kg$ attached to it. The force constant of spring is $10\,N/m$.If at any instant its velocity is $40\,cm/sec$, the displacement will be .... $m$ (where amplitude is $0.5\,m$)
Two pendulums have time periods $T$ and $\frac{{5T}}{4}.$They start $S.H.M.$ at the same time from the mean position. What will be the phase difference between them after the bigger pendulum has complete one oscillation ..... $^o$
The effective spring constant of two spring system as shown in figure will be
A $15 \,g$ ball is shot from a spring gun whose spring has a force constant of $600 \,N/m$. The spring is compressed by $5 \,cm$. The greatest possible horizontal range of the ball for this compression is .... $m$ ($g = 10 \,m/s^2$)