A vertical spring stretches 9.6 cmwhen a 1.3 kgblock is hung from its end. (a) Calculate the spring constant. This block is then displaced an additional 5.0 cmdownward and released from rest. Find the (b) period, (c) frequency, (d) amplitude, and (e) maximum speed of the resulting SHM.

• The spring stretches at x = 9.6 cm or 0.096 m.
• Mass of the block is m = 1.3 kg.
• Additional displacement of the block when it is released from rest is 5.0 cm or 0.050 m.

Using spring force, we can calculate the force constant of the spring. By calculating the frequency of the spring oscillations, we can calculate the period of oscillation of the spring. The oscillations are simple harmonic and start from the rest below the equilibrium position so we can find the amplitude of the spring oscillations. Finally, applying the equation of conservation of energy, we can find the maximum speed of the resulting SHM of the spring.

Formula:

The spring constant of the oscillations, F = kx (i)

The period of oscillations,$T=2\tau \tau \sqrt{\frac{m}{k}}$ (ii)

The potential energy in the string,$PE=-mgx+\frac{1}{2}k{x}^2$ (iii)

The equation of conservation of energy,$TE=PE+\frac{1}{2}m{v}^2$ (iv)

Using equation (i), we get the spring constant as:

$$\begin{gathered} k=\frac{F}{x} \\ =\frac{mg}{x}\left\{\because F=mg\right\} \\ =\frac{1.3\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2}{0.096\mathrm{m}} \\ =\frac{12.74}{0.096} \\ =132.7\mathrm{N}/\mathrm{m} \\ \approx 1.33\times {10}^2\mathrm{N}/\mathrm{m} \end{gathered}$$

Therefore, spring constant is $1.33\times {10}^2\mathrm{N}/\mathrm{m}$.

Using equation (ii) and the given values, the period of oscillations is given as:

$$\begin{gathered} T=2\times 3.14\times \sqrt{\frac{1.3\mathrm{kg}}{132.7\mathrm{N}/\mathrm{m}}} \\ =6.2832\times 0.98977 \\ =0.62\mathrm{s} \end{gathered}$$

Therefore, the period (T) of SHM is

The frequency of oscillations is given by:

$$\begin{gathered} f=\frac{1}{T} \\ =\frac{1}{0.62} \\ =1.6\mathrm{Hz} \end{gathered}$$

Therefore, the frequency of SHM is 1.6 Hz.

The oscillations are simple harmonic and start from the rest 5.0 cm below the equilibrium position so the amplitude of the spring oscillation is 5.0 cm or 0.05 m.

For maximum speedof the resulting SHM, we noted that, the block has maximum speed as it passes the equilibrium position but at initial position it has potential energy,

Potential energy in the spring using equation (iii) is given as:

$P.E_{.0}=-mg{y}_0+\frac{1}{2}k{y_0}^2$

Here

role="math" localid="1657260527567" $$\begin{gathered} y_0=y+y_1 \\ =0.096+0.050 \\ =0.146\mathrm{m} \\ \mathrm{P}.{\mathrm{E}}_{.0}=-(1.3\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 0.146\mathrm{m})+\left(\frac{1}{2}\times 133\mathrm{N}/\mathrm{m}\times {(0.146\mathrm{m})}^2\right) \\ =-1.86+1.4143 \\ =-0.44\mathrm{J} \end{gathered}$$

Now, when block is at the equilibrium, x = 0.096 m. Therefore, potential energy at the equilibrium position using equation (iii) is given as:

$$\begin{gathered} y_0=y+y_1 \\ \mathrm{P}.{\mathrm{E}}_{.}=-(1.3\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 0.096\mathrm{m})+\left(\frac{1}{2}\times 132.7\mathrm{N}/\mathrm{m}\times {(0.096\mathrm{m})}^2\right) \\ =-(1.223)+(0.611) \\ =-0.61\mathrm{J} \end{gathered}$$

By applying law of conservation of energy, we get

$$\begin{gathered} P.E_{.0}=P.E.+\frac{1}{2}m{v}^2 \\ {v}^2=\frac{2(P.E_{.0}-P.E.)}{m} \\ v=\sqrt{\frac{2(P.E_{.0}-P.E.)}{m}} \\ =\sqrt{\frac{2\times \left(\right(-0.44\mathrm{J})-(-0.61\mathrm{J}\left)\right)}{1.3\mathrm{kg}}} \\ =\sqrt{\frac{2\times \left(1658\right)}{1.3}} \\ =0.51\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the maximum speed (v) of SHM is 0.51 m/s.