Q.The displacement–time graph of a particle executing S.H.M. is a cosine curve of period T: the displacement is at its positive maximum at t=0, zero at t=T/4, at its negative maximum (most negative) at t=T/2 (i.e. 2T/4), zero again at t=3T/4, and back at its positive maximum at t=T (i.e. 4T/4), continuing on to t=5T/4. Which of the following statement(s) is/are true? (Note: more than one of the given options may be correct.)
(a) The force is zero at t=43T.
(b) The acceleration is maximum at t=44T.
(c) The velocity is maximum at t=4T.
(d) The P.E. is equal to the K.E. of oscillation at t=2T.
Simple Harmonic Motion: The Natural Rhythm of Things
Imagine a ball placed at the bottom of a perfectly smooth, U-shaped bowl. If you give it a gentle push, what happens? It rolls up one side, slows down, stops for an instant, then rolls back down, past the bottom, up the other side, stops, and returns. Left alone, it keeps doing this — back and forth, back and forth — in a steady, repeating rhythm.
That rhythm is the heart of Simple Harmonic Motion (SHM). It's the most fundamental kind of oscillatory (back-and-forth) motion in physics.
The Intuition: A Restoring Force That Fights Displacement
The key idea is this: the further you push the object from its resting (equilibrium) position, the stronger the force that tries to pull it back.
In the bowl, when the ball is at the bottom (equilibrium), gravity pulls straight down, and the bowl pushes straight up — no sideways force. But when you push the ball up the side, gravity now has a component that pulls it down the slope. The higher up the side you push it, the steeper the slope, and the stronger that pull-back force becomes.
This is a restoring force — it always points toward equilibrium. And crucially, in SHM, this restoring force is directly proportional to the displacement from equilibrium. Double the displacement, double the restoring force.
F=−kx
F is the restoring force.
x is the displacement from equilibrium.
k is a positive constant (the "stiffness" of the system).
The minus sign is crucial: it tells you the force is opposite to the displacement.
The Precise Statement
Simple Harmonic Motion is the motion of an object where the restoring force is directly proportional to the displacement from equilibrium and acts in the opposite direction.
That's it. That single condition — F=−kx — is the entire definition. Everything else (the sine waves, the formulas for period and frequency) follows mathematically from this one law.
What Does This Motion Look Like?
If you track the ball's position over time, you get a beautiful, smooth wave — a sine wave (or cosine wave). It's the same shape as the shadow of a spinning wheel cast on a wall.
The motion has three key descriptors:
Amplitude (A): The maximum displacement from equilibrium. How far you initially pushed the ball up the side of the bowl.
Period (T): The time it takes to complete one full back-and-forth cycle (e.g., from the leftmost point, back to the leftmost point).
Frequency (f): How many cycles happen per second. f=1/T.
Note
A remarkable fact: for a given system (fixed k and fixed mass m), the period and frequency do not depend on the amplitude. A big push and a tiny push take exactly the same time to complete one cycle. This is called isochronism — and it's why pendulums were used to keep time in clocks.
The Mathematical Description (Derived from F=−kx)
Using Newton's second law (F=ma) and the definition of acceleration (a=dt2d2x), the condition F=−kx becomes:
mdt2d2x=−kx
This is a differential equation. Its solution — the position as a function of time — is:
x(t)=Acos(ωt+ϕ)
Where:
ω=mk is the angular frequency (radians per second). It tells you how fast the oscillation is.
ϕ is the phase constant (determines where in the cycle you start measuring time). …
The motion is x=Acos(2πt/T). Force and acceleration ∝−x (max where ∣x∣ is max, zero where x=0); speed is max where x=0. At t=3T/4, x=0 so force is zero (A✓); at t=T, ∣x∣=A so acceleration is max (B✓); at t=T/4, x=0 so velocity is max (C✓). At t=T/2, x=−A (an ex …
Take x(t)=Acosωt with ω=2π/T. In SHM the restoring force and acceleration are proportional to −x (largest at the extremes, zero at the mean), while the speed is largest at the mean position (x=0) and zero at the extremes. Checking the stated instants: force is zero at 3T/4, acceleration is maximum at T, velocity is maximum at T/4 — all true; but at T/2 the particle is at an extreme, so its energy is all potential, making statement (D) false.
Concept and key relations
For x=Acosωt:
v=x˙=−Aωsinωt,a=x¨=−ω2x,F=ma=−mω2x.
So ∣F∣,∣a∣∝∣x∣ (max at extremes, zero at mean) and ∣v∣ is max where x=0. Also PE=21mω2x2 and KE=21mω2(A2−x2); these are equal only when ∣x∣=A/2.
Evaluate each statement
(A) At t=43T: x=Acos(3π/2)=0, so F=−mω2x=0. True. …
Step 1: Model the motion as x(t)=Acos(ωt), ω=2π/T. Then v=−Aωsinωt, a=−ω2x, and F=ma=−mω2x: force and acceleration are largest in magnitude at the extremes (∣x∣=A) and zero at the mean (x=0); speed is largest at the mean and zero at the extremes.
Step 2: At t=43T: x=Acos(3π/2)=0⇒F=0 — statement (a) true.
Step 3: At t=44T=T: x=Acos(2π)=A (an extreme) ⇒∣a∣=ω2A is maximum — statement (b) true. …