Chapter 11, Problem 83GP

To determine

To Check: Out of the vertical oscillation’s periods and horizontal swings period, which one is longer.

Answer to Problem 83GP

The time period of the pendulum oscillations is greater than that of the vibrating oscillations with the factor of . $\sqrt{1+\frac{l_{°}k}{g}}$

Explanation of Solution

Given info:

The length of the unstarched spring is $l_{°}$ ,ad L is the string length with the mass attached at rest.

Formula used:

The time period of the simple harmonic oscillation is given as,

  $T=2\pi \sqrt{\frac{m}{k}}$

Where, m is the mass of the spring, K is the spring constant and T is the time period

Calculation:

When a mass is hanging on the spring, then the spring length after the stretching is L. therefore, the change in length is $L-l_{°}$ . The weight of the mass is equal to the amount of the downward stretch ( $L-l_{°}$ ) that is found from the setting the spring force upward on the mass, such that,

  $\begin{array}{l}k\left(L-l_{°}\right)=mg\\ kL-k{l}_{°}=mg\\ L=l_{°}+\frac{mg}{k}\end{array}$

So, the pendulum length after stretching is $L=l_{°}+\frac{mg}{k}$

Now, the time period for the vertical oscillations is given as,

  $T_{vib}=2\pi \sqrt{\frac{m}{k}}$

And that of the pendulum oscillations is given as,

  $T_{pen}=2\pi \sqrt{\frac{L}{g}}$

Now, divided these time periods with each other such that,

  $\begin{array}{l}\frac{T_{pen}}{T_{vib}}=\frac{2\pi \sqrt{\frac{L}{g}}}{2\pi \sqrt{\frac{m}{k}}}=\frac{\sqrt{\frac{\left(l_{°}+\frac{mg}{k}\right)}{g}}}{\sqrt{\frac{m}{k}}}\\ \frac{T_{pen}}{T_{vib}}=\sqrt{1+\frac{l_{°}k}{g}}\end{array}$

On the bases of the above time-period ratio, it is clear that this period ratio is greater than one, that is

  $\frac{T_{pen}}{T_{vib}}=\sqrt{1+\frac{l_{°}k}{g}}>1$

This implies that the time period for the pendulum oscillations is greater than the period of the vibrating oscillations that is $T_{pen}>T_{vib}$ by the factor of $\sqrt{1+\frac{l_{°}k}{g}}$ .

Conclusion:

Hence, with the factor of $\sqrt{1+\frac{l_{°}k}{g}}$ , the vibrating oscillation’s time period is less than the pendulum oscillation’s time period.