$A$ mass $m$ is suspended by means of two coiled springs which have the same length in the unstretched condition,as shown in the figure. Their force constants are $k_1$ and $k_2$ respectively. When set into vertical vibrations,the period will be:

$A$ 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

When a mass $m$ is attached to a spring,it normally extends by $0.2\, m$. If the mass $m$ is given a slight additional extension and released,what will be its time period?

$A$ mass attached to a spring is free to oscillate,with angular velocity $\omega$,in a horizontal plane without friction or damping. It is pulled to a distance $x_{0}$ and pushed towards the centre with a velocity $v_{0}$ at time $t=0$. Determine the amplitude of the resulting oscillations in terms of the parameters $\omega, x_{0}$ and $v_{0}$. [Hint: Start with the equation $x=A \cos (\omega t+\theta)$ and note that the initial velocity is negative.]

The graph shown was obtained from experimental measurements of the period of oscillations $T$ for different masses $M$ placed in the scale pan on the lower end of the spring balance. The most likely reason for the line not passing through the origin is that the

$A$ $100 \,g$ mass stretches a particular spring by $9.8 \,cm$,when suspended vertically from it. How large a mass must be attached to the spring if the period of vibration is to be $6.28 \,s$?