Solution to Question 20
1. Time Period Calculation
The camera captures $N$ frames for one complete oscillation at a speed of $n$ frames per second. The time period $T$ of the pendulum is: $$T = \frac{N}{n}$$ Given $N = 48$ and $n = 24 \text{ fps}$: $$T = \frac{48}{24} = 2 \text{ s}$$
2. Length of the Actual Pendulum
The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L_{obj}}{g}}$. solving for the actual length $L_{obj}$: $$T^2 = 4\pi^2 \frac{L_{obj}}{g} \implies L_{obj} = \frac{g T^2}{4\pi^2}$$ Substituting $T = \frac{N}{n}$: $$L_{obj} = \frac{g (N/n)^2}{4\pi^2} = \frac{g N^2}{4\pi^2 n^2}$$
3. Magnification and Object Distance
The pendulum acts as the object for the camera lens. Let the distance of the pendulum from the lens be $u$. The size of the object is $L_{obj}$ and the size of the image on the film is $l$. The magnification $m$ is: $$m = \frac{\text{Image Size}}{\text{Object Size}} = \frac{l}{L_{obj}} = \frac{v}{u}$$ Since the object distance is typically much larger than the focal length ($u \gg f$) for such setups, the image forms very close to the focus, so $v \approx f$. $$m \approx \frac{f}{u}$$ Equating the two expressions for magnification: $$\frac{l}{L_{obj}} = \frac{f}{u} \implies u = f \frac{L_{obj}}{l}$$
4. Final Calculation
Substituting the expression for $L_{obj}$: $$u = \frac{f}{l} \left( \frac{g N^2}{4\pi^2 n^2} \right) = \frac{f g N^2}{4\pi^2 n^2 l}$$ Now, substituting the numerical values:
- $f = 50 \text{ mm} = 0.05 \text{ m}$
- $l = 10 \text{ mm} = 0.01 \text{ m}$
- $g \approx \pi^2$ (as per hint)