In the post dated 30^{th} April 2008, a free-response question on simple pendulum (for practice) was given to you. As promised, I give below a model answer along with the question:

*ℓ*is set up

*using a bob of mass*

*m*and a string of negligible mass. The string can withstand a maximum tension of 3

*mg*where

*g*is the acceleration due to gravity. The bob is pulled aside so that the string makes a small angle

*θ*

_{1}with the vertical. On releasing the bob, the pendulum oscillates with period

*T*. Now, answer the following questions:

(a) Calculate the period of oscillation when the length *ℓ* = 1 m, assuming that the acceleration due to gravity at the place is 10 ms^{–2}.

(b) If the bob of the pendulum is now immersed in a liquid of negligible viscosity and of density lower than that of the bob, how will the period be affected? Put a tick mark against the correct option:

Increased ___ Decreased ___ Unchanged ___

Give reason for your answer.

(c) The pendulum bob is taken out from liquid and is oscillated in air itself. Explain what modification you will make to halve the period of oscillation.

(d) The centre of gravity of the bob is raised through a height of 6 cm from its mean position (lowest position) when the bob is in the extreme position during its oscillation. Calculate the speed of the bob when it crosses the mean position.

(e) Calculate the maximum angle through which the string can be displaced from the vertical so that the string will not break when the pendulum oscillates.

(a) The period of oscillation of a simple pendulum of length *ℓ* is given by

*T* = 2π√( *ℓ*/*g*)** **where *g *is the acceleration due to gravity.

Substituting for *ℓ* (=1 m) and *g* (= 10 ms^{–2}), *T = *2π√(0.1) = 1.9869 s.

(b) Within the liquid the bob experiences the force of buoyancy exerted by the liquid so that the effective weight of the bob is *decreased*. The restoring force on the bob is therefore reduced and the *period of oscillation is increased*.

(c) Since the period is directly proportional to the square root of the length of the pendulum, the length is to be reduced to a *quarter *of the original length for making the period *half* the original value.

(d) In the extreme position of the bob, its energy is entirely gravitational potential energy which is *mgh *where *h* is the height through which the bob is raised from its mean position. In the mean position this potential energy is completely converted into kinetic energy so that we have

*mgh *= ½ *mv*^{2} where *v* is the speed of the bob at the mean position.

Thus *v* = √(2*gh*) = √(2×10×0.06) = 1.0954 ms^{–1}.

*mg*(which acts vertically downwards) and the tension in the string (which acts radially towards the centre of the arc of the circle along which the bob moves).

The resultant force acting radially towards the centre (point of suspension) supplies the centripetal force for the circular motion so that we have

*T* – *mg *cos*θ* = *mv*^{2}/*ℓ*

[Note that the radius of the arc of the circle is the length *ℓ* of the pendulum].

Therefore, *T* = *mg *cos*θ* + *mv*^{2}/*ℓ*

In the mean position we have *θ* = 0 so that the above equation becomes

*T* = *mg *+ *mv*^{2}/*ℓ*

Since the maximum tension that the string can withstand is 3*mg*, we have

3*mg *= *mg *+ *mv*^{2}/*ℓ*, from which *mv*^{2}/*ℓ* = 2*mg.*

Therefore, *v ^{2}* = 2

*gℓ*.

This means that in the extreme position the bob is raised through a height *ℓ* with respect to its mean position [Since *v* = √(2*gh*) with usual notations].

The angle which the string makes with the vertical is therefore 90º in the limiting case. [In other words this is the maximum angle through which the string can be displaced from the vertical so that it will not break when the pendulum oscillates].

* * * * * * * * * * * * * * * * * * * *

[Suppose the string would break when the tension is 2*mg*.

You will then have 2*mg *= *mg *+ *mv*^{2}/*ℓ*, from which *mv*^{2}/*ℓ* = m*g.*

Therefore, *v ^{2}* =

*gℓ*. Since

*v*

^{2}= 2

*gh*,

*h*=

*ℓ*/2.

But *h* = *ℓ*(1– cos*θ*_{1}) where *θ*_{1} is the maximum angle through which the string can be displaced from the vertical. Therefore, (1– cos*θ*_{1}) = ½ from which *θ*_{1} = 60º].

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