Name: Condition for Resonance
Uploaded: 2019-07-19
Duration: 5 min 17 s
Description: Do you know you can shatter the glass goblet without touching it? Yes, it is possible when the sound of a specific frequency is produced. Let's find out the...

Question

A pendulum is constructed as a light thin-walled sphere of radius $R$ , filled up with water and suspended at the point $O$ from a light rigid rod. The distance between the point $O$ and the centre of the sphere is equal to $l .$ How many times will the small oscillations of such a pendulum change, after the water freezes? The viscosity of water and the change of its volume on freezing are to be neglected.

Accepted Answer

The system will consist of the light rod and the frozen water (as the water got frozen) and the hollow sphere form a compound (physical) pendulum to an excellent approximation, as the whole system can be taken to be rigid. For such systems, the time period is stated by,

$\Rightarrow τ_{R e} + I α = 0 \Rightarrow m g l θ + (\frac{2}{5} m R^{2} + m l^{2}) \ddot{θ} = 0 \Rightarrow ω = \sqrt{\frac{g l}{\frac{2}{5} R^{2} + l^{2}}} T_{1} = \frac{2 π}{ω} = 2 π \sqrt{\frac{\frac{2}{5} R^{2} + l^{2}}{g l}} . . . (1)$

$T_{1} = 2 π \sqrt{\frac{l}{g}} \sqrt{1 + \frac{k^{2}}{l^{2}}}$ , where the radius of gyration of the sphere is $k^{2} = \frac{2}{5} R^{2}$ .

The situation changes when the water gets unfrozen. When we neglect dissipative forces (viscosity), we are dealing with ideal fluids. Such fluids will instantaneously respond to (unbalanced) internal stresses. Assume that the sphere with liquid water actually executes small rigid oscillations. The portion of the fluid above the centre of the sphere will have greater acceleration than that below the centre. This is because the linear acceleration of any element is, in this case, equal to the angular acceleration of the element multiplied by its distance from the centre of suspension (since we are considering small oscillations). Then, as is obvious in a frame moving with the centre of mass, we will have an unbalanced couple (not negated by any pseudo forces) which will cause the fluid to move rotationally, to destroy differences in acceleration. Hence, in case of ideal fluids, the pendulum must move such that the elements of the fluid, all undergo the same acceleration. This implies, we have a simple (mathematical) pendulum with the time period,

$T_{0} = 2 π \sqrt{\frac{l}{g}} . . . (2)$

Form equations $(1) & (2)$ , $T_{1} = T_{0} \sqrt{1 + \frac{2}{5} {(\frac{R}{l})}^{2}}$

(One expects that a liquid with very small viscosity to have a time period close to $T_{0}$ . On the other hand, one with high viscosity will have a time period closer to $T_{1}$ )

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