Q.(a) Define simple harmonic motion. Show that the motion of (point) projection of a particle performing uniform circular motion, on any diameter, is simple harmonic.
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Start your 14-day free trial to unlock the full solution →(a) SHM is motion with acceleration a = −ω^2 x; the foot of the perpendicular from a particle in uniform circular motion, dropped onto a diameter, obeys exactly this equation, proving it is SHM. (b) Setting v_max = (1/2) a_max gives ω = 2 rad/s, so T = π s.
(a) Definition of SHM: Simple harmonic motion is a periodic oscillatory motion in which the restoring force (and hence the acceleration) on the particle is always directed towards a fixed mean (equilibrium) position and is directly proportional to the particle's displacement from that mean position:
a = −ω^2 x
where ω is a constant (angular frequency) and x is the displacement from the mean position. The negative sign shows the acceleration/force always opposes the displacement (restoring in nature).
Proof that the projection of uniform circular motion is SHM:
Consider a particle P moving with uniform angular velocity ω on a circle of radius A, centred at O. Let N be the foot of the perpendicular dropped from P onto a fixed diameter (say, the x-axis) — N is the 'projection' of P on that diameter.
At time t, if P started at angle 0 on the x-axis, the angle swept is θ = ωt, and P's coordinates are (A cos ωt, A sin ωt). The projection N on the x-axis has position:
x(t) = A cos ωt
Differentiate twice with respect to time to get the acceleration of N:
v = dx/dt = −Aω sin ωt
a = dv/dt = −Aω^2 cos ωt = −ω^2 (A cos ωt) = −ω^2 x
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