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 dy/dt = A ωcos 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 a0 at the end of 100 oscillations. When the oscillator completes 200 oscillations, its amplitude must be
(a) a0 /2
(b) a0 / 6
(c) a0 /12
(d) a0 /4
(e) a0 /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 = A0 exp(– bt) where A0 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
a0/3 = a0 exp(– bt) and
a = a0 exp(– 2bt) where ‘a’ is the amplitude after 200 oscillations (at time 2t)
From the first equation, exp(– bt) = 1/3. Substituting this in the second equation,
a = a0×(1/3)2 = a0/9.
(3) For a simple pendulum, the graph between T2 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
T2 = (4π2/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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