Question

A pan with a set of weights is attached to a light spring. The period of vertical oscillation is $0.5s$. When some additional weights are put in the pan, then the period of oscillations increases by $\mathbf{0}.\mathbf{1s}$. The extension caused by the additional weight is

Answer 1

We will use the time period oscillation formula, to find the displacement distance done by the spring due to the weights added. Form a time period equation, where the total time is period of vertical oscillation plus the additional increase in time due to additional weight on the pan and similarly add both the weights (set of weights plus additional weights). After these squares both sides of and find the displacement distance.

Using the time period formula for the oscillation:
$T=2\pi \sqrt{\dfrac{m}{k}}$
And the time period for extension when under influence of gravity is:
$T=2\pi \sqrt{\dfrac{l}{g}}$
where $T$ is the time period of the oscillation, $m$ is the mass of the substance, $k$ is the spring constant in unit of $\text{Newton/meter}$, $l$ is the displacement of the spring from its equilibrium and $g$ is the gravitational force in $m{{s}^{-2}}$.

Complete step by step solution:
Now to find the displacement of the pan when additional weights are added, first the time period of the oscillation without the additional weights is:
$T=0.5s$
And the time period formula in terms of mass and spring constant is:
$T=2\pi \sqrt{\dfrac{m}{k}}$
Placing the values of the time period and forming relationship between them we get:
$0.5=2\pi \sqrt{\dfrac{m}{k}}$
Squaring both the sides of the LHS and RHS,
${{\left( 0.5 \right)}^{2}}={{\left( 2\pi \sqrt{\dfrac{m}{k}} \right)}^{2}}$
$\Rightarrow 0.25=4\left( {{\pi }^{2}} \right)\dfrac{m}{k}$
Converted $0.25$ to $\dfrac{1}{4}$, we get:
$\Rightarrow \dfrac{1}{4\times 4}=\left( {{\pi }^{2}} \right)\dfrac{m}{k}$
$\Rightarrow \dfrac{m}{k}=\dfrac{1}{16{{\pi }^{2}}}$ .…(i)
Now adding the time period with the additional time along with the additional weight $\left( {{m}_{a}} \right)$ on the pan, we get:
$T+0.1s=2\pi \sqrt{\dfrac{m+{{m}_{a}}}{k}}$
Placing the value of time period $T=0.5s$, in the above formula, we get the equation as:
$T+0.1s=2\pi \sqrt{\dfrac{m+{{m}_{a}}}{k}}$
$\Rightarrow 0.5s+0.1s=2\pi \sqrt{\dfrac{m+{{m}_{a}}}{k}}$
$\Rightarrow 0.6s=2\pi \sqrt{\dfrac{m+{{m}_{a}}}{k}}$
Squaring both the sides of the LHS and RHS,
$\Rightarrow {{\left( 0.6s \right)}^{2}}={{\left( 2\pi \sqrt{\dfrac{m+{{m}_{a}}}{k}} \right)}^{2}}$
$\Rightarrow 0.36s=4{{\pi }^{2}}\left( \dfrac{m+{{m}_{a}}}{k} \right)$
Separating the fractions for masses into two one for the mass and another for the additional mass:
$\Rightarrow \dfrac{0.36s}{4{{\pi }^{2}}}=\left( \dfrac{m}{k}+\dfrac{{{m}_{a}}}{k} \right)$
Placing the value of $\dfrac{m}{k}=\dfrac{1}{16{{\pi }^{2}}}$ in the above equation:
$\Rightarrow \dfrac{0.36s}{4{{\pi }^{2}}}=\left( \dfrac{1}{16{{\pi }^{2}}}+\dfrac{{{m}_{a}}}{k} \right)$
$\Rightarrow \dfrac{0.44s}{16{{\pi }^{2}}}=\dfrac{{{m}_{a}}}{k}$
Now both the formulas for time period when in terms of spring constant or gravity when brought together and equated to find the equivalent term, we get:
$T=2\pi \sqrt{\dfrac{m}{k}}=2\pi \sqrt{\dfrac{l}{g}}$
$\Rightarrow 2\pi \sqrt{\dfrac{m}{k}}=2\pi \sqrt{\dfrac{l}{g}}$
$\Rightarrow \sqrt{\dfrac{m}{k}}=\sqrt{\dfrac{l}{g}}$
$\Rightarrow \dfrac{m}{k}=\dfrac{l}{g}$, (Here mass $m$ is equivalent to ${{m}_{a}}$)
Hence, the equation $\dfrac{0.44s}{16{{\pi }^{2}}}=\dfrac{{{m}_{a}}}{k}$ can also be written as:
$\dfrac{0.44s}{16{{\pi }^{2}}}=\dfrac{l}{g}$
Placing the value of gravity as $g=9.8m{{s}^{-2}}$, we get the value of $l$ or extension caused by additional weight is:
$\Rightarrow \dfrac{0.44s}{16{{\pi }^{2}}}=\dfrac{l}{g}$
$\Rightarrow \dfrac{0.44s}{16{{\pi }^{2}}}=\dfrac{l}{9.8}$
$\Rightarrow l=\dfrac{0.44s}{16{{\pi }^{2}}}\times 9.8$
$\Rightarrow l=\left( 0.0279\times 9.8 \right)m$
$\Rightarrow l=2.73cm$
Therefore, the extension length when additional weight is applied on the pan is $2.73cm$.

**Note : **A point should be always remembered that for this question we need both the formulas of time period one in respect to mass and other with respect to length. Although an additional mass and mass are talked about in the question but there is no data given for them that doesn’t mean we only need the formula:
$T=2\pi \sqrt{\dfrac{l}{g}}$
Although the above formula has all the variables known but still we need to equate both the formulas to make sure that the time period and its extension is based on the mass of the additional weight only as $T=2\pi \sqrt{\dfrac{l}{g}}$ only talks about the weight of the pan and neglecting the weight of the additional mass on the pan.