Question

A small block oscillates back and forth on a smooth concave surface of radius R (figure 12-E17). Find the time period of small oscillation

Answer 1

Let the mass of the small block be m,

Refer to the attachment.

The normal force on the block = mg.cosθ and the tangential force on the block is mgsinθ.

Thus, Acceleration of the block = a

= mgsinθ/m

= gsinθ

Now, block is very small, and it makes small oscillations as given in question, which signifies that the angular displacement will also be small. Now, when θ is small, then sinθ ≈ θ

Thus, acceleration a = gsinθ

= gθ

Let the linear displacement covered by the block from the lowest position of curve be x. Then θ = x/R

∴  a =gx/R =(g/R)x

Since, g/R is constant, hence the acceleration is proportional to the displacement. Thus motion is simple harmonic motion.

Comparing it with a = w²x.

We will get, w = √(g/R)

Thus, The time period of the oscillation = T

∴ $T =\frac{2\pi }{\omega}$

∴ $T = 2\pi \frac{R}{g}$

Hence, the time period of the small oscillations is  $T = 2\pi \frac{R}{g}$.

Hope it helps.

Answer 2

Answer:

The normal force on the block = mg.cosθ and the tangential force on the block is mgsinθ.

Thus, Acceleration of the block = a

= mgsinθ/m

= gsinθ

Now, block is very small, and it makes small oscillations as given in question, which signifies that the angular displacement will also be small. Now, when θ is small, then sinθ ≈ θ

Thus, acceleration a = gsinθ

= gθ

Let the linear displacement covered by the block from the lowest position of curve be x. Then θ = x/R

∴  a =gx/R =(g/R)x

Since, g/R is constant, hence the acceleration is proportional to the displacement. Thus motion is simple harmonic motion.

Comparing it with a = w²x.

We will get, w = √(g/R)