Equations to be remembered in the section ‘oscillations’ were discussed in the post dated 17labe l ‘oscillation’ below this post. Today we will discuss a few multiple choice practice questions involving simple pendulum:

^{th}April 2008. Some multiple choice practice questions in this section were discussed in the post dated 22^{nd}April 2008, followed by a free response practice question in the post dated 2^{nd}May 2008. You can access all these posts by clicking on the
(1) A simple pendulum arranged inside a train has period of oscillation

*T*when the train is at rest. When the train moves along straight horizontal rails with uniform*acceleration*of*x*ms^{–2}, the period of the pendulum is (assuming*g*= 10 ms^{–2})
(a)

*T*×*g*^{1/2}/(*g+x*)^{1/2}^{}
(b)

*T*×*g*^{1/4}/(*g+x*)^{1/4 }
(c)

*T*×*g*^{1/4}/(*g*^{2}*+x*^{2})^{1/4}
(d)

*T*×*g*^{1/2}/(*g*^{2}*+x*^{2})^{1/2}
(e)

*T*×*g*^{1/2}/(*g*^{2}*+x*^{2})^{1/4}
When the train is at rest the period of oscillation

*T*of the pendulum is given by*T*= 2π (

*ℓ*/

*g*)

^{1/2}………….(i)

where

*ℓ*is the length of the pendulum and*g*is the acceleration due to gravity.
When the train moves forward with acceleration

*x*ms^{–2}, an inertial backward force acts on the bob of the pendulum and supplies a backward acceleration of*x*ms^{–2}(fig.). The resultant acceleration of the bob is (*g*^{2}*+x*^{2})^{1/2}. The period*T*_{1}*of the pendulum is now given by**T*

_{1}=2π[

*ℓ*/(

*g*

^{2}

*+x*

^{2})

^{1/2}]

^{1/2}……..(ii)

Dividing eqn (ii) by eqn (i) we have

*T*

_{1}/

*T =*

*g*

^{1/2}/(

*g*

^{2}

*+x*

^{2})

^{1/4}from which

*T*

_{1}=

*T*×

*g*

^{1/2}/(

*g*

^{2}

*+x*

^{2})

^{1/4}

The correct option is (e)

[Suppose the train is moving with uniform

*velocity*of*x*ms^{–1}. What will be the period? No doubt,*T*itself since you cannot distinguish between state of rest and uniform motion].
(2) Two simple pendulums A and B have periods 2.1 s and 2 s respectively. They start oscillating at the same time

*in phase*. They will be in phase instantly at the end of
(a) 42 s

(b) 40 s

(c) 22 s

(d) 21 s

(e) 20 s

Suppose the pendulums are in phase instantly at the end of

*n*oscillations of pendulum A. Pendulum B should then execute*n*+1 oscillations. Therefore we have*n*×2.1 = (

*n*+1)×2

Therefore 0.1

*n =*2 so that n = 20
The time elapsed is therefore 20×2.1 = 42 s [Option (a)].

[The difference between the periods of the two pendulums is 0.1 s. So you require 20 oscillations of pendulum A to obtain a time difference equal to one period (2 s) of pendulum B so that the two pendulums will be instantly in phase].

(3) The bob of a simple pendulum is a hollow metal sphere filled with water. There is a small hole at the bottom of this hollow sphere and water drains out through the hole as the pendulum oscillates. After completely filling the bob with water, if this pendulum is made to oscillate for a long time, its period of oscillation will

(a) increase first and will reach a final constant value

(b) decrease first, reach a minimum value, then increase and will finally settle at the initial value

(c) increase first, reach a maximum value, then decrease and will finally settle at the initial value

(d) decrease first and will reach a final constant value

(e) remain unchanged

The correct option is (c). Initially the centre of gravity of the spherical bob is at its centre since it is completely filled with water. When the water flows out, the centre of gravity of the bob moves gradually downwards, reaches a minimum level and then moves up. When the water is fully drained out, the centre of gravity of the bob once again reaches the centre of the bob and remains there.

The length of the pendulum is the distance between the centre of gravity of the bob and the point of suspension. Therefore, the length of the pendulum gets increased initially, becomes a maximum, then gets decreased and finally settles at the initial value. Therefore, the period of the pendulum will increase first, reach a maximum value, then decrease and will finally settle at the initial value.

(4) A simple pendulum is taken to a location where the acceleration due to gravity is decreased by 0.1 %. If the period of oscillation is to be unaltered

(a) the mass of the bob is to be increased by 0.1 %

(b) the mass of the bob is to be decreased by 0.1 %

(c) the length of the pendulum is to be decreased by 0.316 %

(d) the length of the pendulum is to be decreased by 0.2 %

(e) the length of the pendulum is to be decreased by 0.1 %

Since the period of oscillation

*T*is given by*T*= 2π√(*ℓ*/*g*), the ratio*ℓ*/*g*should be unaltered. Therefore, the length*ℓ*should be decreased by 0.1 %
The following questions are for AP Physics C aspirants only:

(5) The period of a simple pendulum is decreased by 0.02 s when the length of the pendulum is decreased by 1 cm. The original length of the pendulum is nearly (

*g =*10 ms^{–2})

(a) 0.25 m

(b) 0.5 m

(c) 1 m

(d) 1.01 m

(e) 2 m

The period of oscillation

*T*is given by*T*= 2π√(*ℓ*/*g*).
Therefore, ∆

*T/T =*½ ∆*ℓ/ℓ*– ½ ∆*g/g*
[Here ∆

*T*,*∆**ℓ*, and*∆**g*represent the increments in the period, length and acceleration due to gravity respectively. We have written the above equation by taking the logarithm of the expression for period and then differentiating it]
The

*increment*in*T*is – 0.02 s and the*increment*in*ℓ*is – 0.01 m (negative signs are because*T*and*ℓ*are*decreased*). There is no change in*g*.
Therefore we have – 0.02/

*T =*– ½ × 0.01/*ℓ*
This gives

*T*= 4*ℓ*.
But

*T*= 2π√(*ℓ*/*g*) so that 2π√(*ℓ*/10) = 4*ℓ*
Or, π

^{2}/10 = 4*ℓ*
Since π

^{2}is nearly equal to 10, we obtain*ℓ*= 0.25 m, nearly.
The simple gravity pendulum is an idealized mathematical model of a pendulum. This is a weight (or bob) on the end of a massless cord suspended from a pivot, without friction. When given an initial push, it will swing back and forth at a constant amplitude. Real pendulums are subject to friction and air drag, so the amplitude of their swings declines.. I am a college sophomore with a dual major in Physics and Mathematics @ University of California, Santa Barbara. By the way, i came across these excellent physics flash cards. Its also a great initiative by the FunnelBrain team. Amazing!!!

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