Question 306: Oscillations

Question

A mass $$M$$ is suspended from a light spring. An additional mass $$M_1$$ added extends the spring further by a distance $$x$$. Now, the combined mass will oscillate on the spring with period $$T=$$

Accepted Answer

Time period of oscillation when mass $$M$$ hangs by spring of spring constant $$k$$ is given by,
$$T=2 \pi \sqrt{M / k}$$
From Hooke's law, $$k x=m g$$
$$\Rightarrow \quad \frac{1}{k}=\frac{x}{m g}$$
When mass $$M_1$$ is added to the system
$$T^{\prime}=2 \pi \sqrt{\frac{\left(M_1+M\right)}{k}}=2 \pi \sqrt{\frac{x\left(M_1+M\right)}{m g}}$$

Answer

$$2 \pi\left[\left(\frac{M_1 g}{x\left(M+M_1\right)}\right)\right]^{\frac{1}{2}}$$

Answer

$$2 \pi\left[\frac{\left(M+M_1\right) x}{M_1 g}\right]^{\frac{1}{2}}$$

Answer

$$\left(\frac{\pi}{2}\right)\left[\left(\frac{M_1 g}{x\left(M+M_1\right)}\right)\right]^{\frac{1}{2}}$$

Answer

$$2 \pi\left[\left(\frac{M+M_1}{M_1 g x}\right)\right]^{\frac{1}{2}}$$

Oscillations

A mass $M is suspended from a light spring. An additional mass M_1 added extends the spring further by a distance x. Now, the combined mass will oscillate on the spring with period T=

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