Question

Two springs have force constants ${{k}_{1}}$ and ${{k}_{2}}$. They are attached to mass $\text{m}$ and two fixed supports.
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If the surface is frictionless, find frequency of oscillations and spring factors of the combination.

Answer 1

Firstly, separate values of restoring forces ${{\text{F}}_{1}}$ and ${{\text{F}}_{2}}$ for springs $1$ and $2$ is calculated and then total restoring force is obtained.
Comparing it with the standard equation gives the value of force constant $k$ and then frequency is calculated.

Formula used $\text{F = }-ky$, $f=\dfrac{1}{2\pi }\sqrt{\dfrac{k}{m}}$

Complete Step by step solution
When mass $\text{m}$ is displaced through some distance, say $y$ towards spring on left, it will get compressed and the other one on the right will get elongated.
But the restoring forces ${{\text{F}}_{1}}$ and ${{\text{F}}_{2}}$ developed in the two springs will be towards right that is in the same direction.
As ${{k}_{1}}$ and ${{k}_{2}}$ are force constants of the two springs, therefore
${{\text{F}}_{1}}=-{{k}_{1}}y$ and ${{\text{F}}_{2}}=-{{k}_{2}}y$
Total restoring force

& \text{F = }{{\text{F}}_{1}}\text{+ }{{\text{F}}_{2}} \\\ & \text{F = }-{{k}_{1}}\text{y}-{{k}_{2}}\text{y} \\\ \end{aligned}$$ $$\text{F = }-\left( {{k}_{1}}+{{k}_{2}} \right)\text{y}$$ …..(1) In the given arrangement, $$\text{F = }-ky$$ …..(2) From equation (1) and (2), we get $$k={{k}_{1}}+{{k}_{2}}$$ Hence, frequency of oscillation of given spring system can be calculated as: $$\begin{aligned} & v=\dfrac{1}{2\pi }\sqrt{\dfrac{k}{m}} \\\ & v=\dfrac{1}{2\pi }\sqrt{\dfrac{{{k}_{1}}+{{k}_{2}}}{m}} \\\ \end{aligned}$$ **Note** Knowledge of concepts regarding motion of loaded spring should be there. Restoring force, which acts on a body to bring back its equilibrium position. The formula used is $$\text{F = }-ky$$, where $$\text{F}$$is restoring force, $$k$$ is force constant and $$y$$ is displacement.