Find the time period of mass $M$ when displaced from its equilibrium position and then released for the system shown in figure.
Here, for calculation of time period we have neglected acceleration of gravity. Because, it is constant at all places and doesn't affect on resultant restoring force.
Let in the equilibrium position, the spring has extended by an amount $x_{0}$. Let the mass be pulled through a distance $x$ and then released. But, string is inextensible, hence the spring alone will contribute the total extension $x+x=2 x$. So net extension in the spring will be $2 x+x_{0}$.
When the extension of spring is $x_{0},(\mathrm{M}$ is not suspended) the restoring force in spring $\mathrm{F}=2 k x_{0}, \ldots$ (1) $\left[\because \mathrm{F}=\mathrm{T}+\mathrm{T}\right.$ and $\left.\mathrm{T}=k x_{0}\right]$
When mass is suspended, the net extension in spring $2 x+x_{0}$ and restoring force in spring, $\mathrm{F}^{\prime}=2 k\left(2 x+x_{0}\right)=4 k x+2 k x_{0} \quad \ldots$ (2)
Restoring force on system,
$f^{\prime} =-\left(\mathrm{F}^{\prime}-\mathrm{F}\right)$
$=\mathrm{F}-\mathrm{F}^{\prime}$
$=2 k x_{0}-4 k x-2 k x_{0} \quad \text { [From equ. (1) and (2)] }$
$=-4 k x$
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A block with mass $M$ is connected by a massless spring with stiffiess constant $k$ to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude $A$ about an equilibrium position $x_0$. Consider two cases: ($i$) when the block is at $x_0$; and ($ii$) when the block is at $x=x_0+A$. In both the cases, a perticle with mass $m$ is placed on the mass $M$ ?
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($C$) The total energy decreases in both the cases
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