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Q.Define simple harmonic motion. Show that the motion of (point) projection of a particle performing uniform circular motion, on any diameter, is simple harmonic. On an average a human heart is found to beat 75 times in a minute. Calculate its frequency and period.

Andhra Pradesh BieapBIEAP Intermediate Board (1st Year) 2019Subjective· 8mImportance★★★★★
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SHM is oscillatory motion with a restoring force proportional to (and opposite to) displacement from a mean position; the shadow/projection of uniform circular motion on a diameter is a classic example of SHM. A heart beating 75 times a minute has frequency 1.25 Hz and period 0.8 s.

Definition of simple harmonic motion

Simple harmonic motion (SHM) is a type of periodic (oscillatory) motion in which the restoring force acting on the particle is always directed towards a fixed point (the mean/equilibrium position) and its magnitude is directly proportional to the displacement of the particle from that mean position:

F=−kxF = -kx

where xx is the displacement from the mean position and kk is a positive constant (the force constant). The negative sign shows the force always opposes the displacement, pulling the particle back towards the mean position. Correspondingly, the acceleration is a=−ω2xa = -\omega^2 x, where ω\omega is the angular frequency of the SHM.

Proof: projection of uniform circular motion is SHM

Consider a particle P moving with constant angular speed ω\omega in a circle of radius AA, centred at O, in the XY-plane. At time t=0t = 0, suppose P is on the X-axis. At time tt, the radius OP makes an angle θ=ωt\theta = \omega t with the X-axis (measuring from some reference).

Let N be the foot of the perpendicular from P onto the Y-axis (i.e., N is the projection, or 'shadow', of P onto the Y-axis, or equivalently consider projecting onto any fixed diameter — here we take the Y-axis for concreteness).

The position of N (i.e., the projection's displacement along the Y-axis) is:

y=Asin⁡θ=Asin⁡(ωt)y = A\sin\theta = A\sin(\omega t)

Differentiating with respect to time to get the velocity of the projection:

dydt=Aωcos⁡(ωt)\frac{dy}{dt} = A\omega\cos(\omega t)

Differentiating again to get the acceleration:

d2ydt2=−Aω2sin⁡(ωt)=−ω2y\frac{d^2y}{dt^2} = -A\omega^2\sin(\omega t) = -\omega^2 y

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