The following questions (MCQ) appeared in Kerala Engineering Entrance (2007) question paper:

**(1) A particle executes S.H.M. with a time period of 16 s. At time t = 2 s, the particle crosses the mean position while at t = 4 s, its velocity is 4 ms**^{–1}. The amplitude of motion in metre is

**(a) √2 π**

**(b) 16√2 π **

**(c) 24√2 π **

**(d) 4/π **

**(e) (32√2)/π **** **

If we start *reckoning time from the instant the particle crosses the mean position *(*y* =0 when *t* =0), the equation of the simple harmonic motion can be written as

*y *= *A *sin *ωt*, where *A* is the amplitude and *ω* is the angular frequency.

In this case, at the instant *t* = 2 s, we have velocity d*y*/d*t* = *A** ω*cos* ωt* = 4** **ms^{–1}.** **

Since the period *T* = 16 s, *ω*= 2π/*T *= π/8 so that the above equation becomes

*A*×(π/8) ×cos [(π/8)×2] = 4

Or, *A*×(π/8)×(1/√2) = 4, from which *A* = **(32√2)/π **** **

**(2) In damped oscillations, the amplitude of oscillations is reduced to one-third of its initial value a _{0} at the end of 100 oscillations. When the oscillator completes 200 oscillations, its amplitude must be**

**(a) a _{0} /2**

**(b) a _{0} / 6**

**(c) a _{0} /12**

**(d) a _{0} /4**

**(e) a _{0} /9**

In the case of damped oscillations, the amplitude decrease *exponentially* with time** **so that at time *t*, the amplitude (A) is given by

** A = A_{0} exp(– bt) **where A

_{0}is the initial amplitude (at

*t*= 0) and

*b*is the damping constant.

If the time required for 100 oscillations is ‘t’, the time required for 200 oscillations will be 2t. Therefore we have

a_{0}/3 = a_{0} exp(– *b*t) and

a = a_{0} exp(– 2*b*t) where ‘a’ is the amplitude after 200 oscillations (at time 2t)

From the first equation, exp(– *b*t) = 1/3. Substituting this in the second equation,

a = a_{0}×(1/3)^{2}_{ }= a_{0}/9.

**(3) For a simple pendulum, the graph between T ^{2} and L is**

**(a) a straight line passing through the origin**

**(b) parabola**

**(c) circle**

**(d) ellipse**

**(e) hyperbola **

The period (T) is given by T = 2 π√(L/g) with usual notations so that

T^{2} = (4π^{2}/g)L

This is in the form y = mx, which represents a straight line passing through the origin. So, the correct option is (a).

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