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Question

A mass m is gently placed on the end of a freely hanging spring. The mass then falls 33 cm before it stops and begins to rise. What is the frequency of the oscillation?

Solution

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Answered 2 years ago
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Given values:\text{\underline{\textbf{Given values:}}}

H=33 cm=0.33 mH=33 \ \text{cm}=0.33 \ \text{m}

g=9.81 ms2g=9.81 \ \dfrac{\text{m}}{\text{s}^2}

If we assume that the equilibrium position of the string in:

x=0andy=Hx=0 \quad \quad \text{and} \quad \quad y=H

and streceted position of the string in:

x=Handy=0x=H \quad \quad \text{and} \quad \quad y=0

We can start from energy of conservation relation who is given as:

Etotal=Ep+Ek.E_{total}=E_{p}+E_{k}.

After plugging values of EtotalE_{total}, EpE_{p} and EkE_{k} in above relation, we will get relation for amplitude:

12kA2=12kx2+12mυ2\begin{align*} \frac{1}{2} kA^2 &=\frac{1}{2} k x^2 +\frac{1}{2} m \upsilon^2 \\ \end{align*}

Since the kinetic energy is equal zero and the potential energy at equilibrium position is equal to the potential energy a stretched position, we get:

12mυE2+mgyE+12kxE2=12mυs2+mgys+12kxs2\frac{1}{2} m \upsilon_{E}^2+mgy_{E}+\frac{1}{2}kx_{E}^2=\frac{1}{2} m \upsilon_{s}^2+mgy_{s}+\frac{1}{2}kx_{s}^2

After rearrange above relation:

0+mgyE+0=0+0+12kxs20+mgy_{E}+0=0+0+\frac{1}{2}kx_{s}^2

As the yE=Hy_{E}=H and xs2=H2x_{s}^2=H^2, previous relation becomes:

mgH=12kH2mgH=\frac{1}{2}kH^2

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