Question

A simple pendulum is attached to block which slides without friction down an inclined plane(ABC) having an angle of inclination $\alpha$ as shown while the block is sliding down the pendulum oscillates in such a way that at its mean position the direction of the string is$?$

• A. At angle $\alpha$ to the perpendicular to the inclined plane AC
• B. Parallel to the inclined plane AC
• C. Vertically downwards
• D. Perpendicular to the inclined plane AC

Answer 1

The correct option is D Perpendicular to the inclined plane AC

Let us assume that the mean position of the pendulum is at an angle $\theta$ with the vertical.

The free body diagram of the mass is shown in the figure at mean position.

At equilibrium position, the net force on the pendulum must be zero.

Net horizontal force = $T\text{cos}(90-\theta )-mg\text{sin}\left(\alpha \right)\text{cos}\left(\alpha \right)=0$

$\Rightarrow \text{sin}\theta =\frac{mg\text{sin}\left(\alpha \right)\text{cos}\left(\alpha \right)}{T}$ ----------(1)

Net vertical force = $T\text{sin}(90-\theta )+mg{\text{sin}}^2\left(\alpha \right)-mg=0$

$\Rightarrow \text{cos}\theta =\frac{mg{\text{cos}}^2\left(\alpha \right)}{T}$ ----------(2)

Dividing (1) and (2), we have

$\text{tan}\theta =\text{tan}\alpha \Rightarrow \theta =\alpha$

Thus, the mean position of the pendulum is perpendicular to the inclined plane AC.