## Related questions with answers

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

Verified$\text{\underline{\textbf{Given values:}}}$

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

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

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

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

and streceted position of the string in:

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

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

$E_{total}=E_{p}+E_{k}.$

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

$\begin{align*} \frac{1}{2} kA^2 &=\frac{1}{2} k x^2 +\frac{1}{2} m \upsilon^2 \\ \end{align*}$

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:

$\frac{1}{2} m \upsilon_{E}^2+mgy_{E}+\frac{1}{2}kx_{E}^2=\frac{1}{2} m \upsilon_{s}^2+mgy_{s}+\frac{1}{2}kx_{s}^2$

After rearrange above relation:

$0+mgy_{E}+0=0+0+\frac{1}{2}kx_{s}^2$

As the $y_{E}=H$ and $x_{s}^2=H^2$, previous relation becomes:

$mgH=\frac{1}{2}kH^2$

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