Question

The physical pendulum in Fig. 15-62 has two possible pivot points A and B. Point A has a fixed position but B is adjustable along the length of the pendulum as indicated by the scaling. When suspended from A, the pendulum has a period of$\mathbf{T}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{s}$. The pendulum is then suspended from B, which is moved until the pendulum again has that period. What is the distance L between A and B?

Short answer

Length between A and B is 0.804 m

Step-by-step solution

The given data

The period of oscillations, $T=1.8\sec$.

Understanding the concept of SHM

The pendulum is suspended from B, but according to the problem, there is no change in the period when it is suspended from A. Sousing the basic formula of the period in terms of length and acceleration due to gravity, we can find the length between A and B.

Formula:

The period of an oscillation, $T=2\pi \sqrt{\frac{L}{g}}$ (i)

Calculation of length between A and B

Though the pendulum is suspended from B, but according to the problem, there is no change in the time period. We can treat the physical pendulum as a simple pendulum with a length equal to the length from the pivot point to its center of mass.

Using equation (i) and the given value of the period, we can get the length of the pendulum between A and B is given as:

$$\begin{gathered} L=\frac{T^{2}g}{4{\pi }^{\mathit{2}}} \\ =\frac{{(1.8s)}^2\times (9.8\mathrm{m}/{\mathrm{s}}^2)}{4{\pi }^{\mathit{2}}} \\ =0.804\mathrm{m} \end{gathered}$$

Hence, the value of the length is 0.804 m.