Q.A particle executing SHM crosses points A and B with the same velocity. Having taken 3 s in passing from A to B, it returns to B after another 3 s. The time period is a) 15 s b) 6 s c) 12 s d) 9 s
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).
From ω, you get the period: T=ω2π=2πkm.
Watch out
Do not confuse angular frequency ω (rad/s) with ordinary frequency f (Hz). They are related by ω=2πf. Many exam errors come from mixing these up.
Real-World Examples
SHM is an idealization — a perfect model. But many real systems approximate it beautifully:
A mass on a spring (horizontal or vertical) — the classic textbook example.
A simple pendulum — but only for small angles (less than about 15∘). For large swings, the restoring force is no longer proportional to displacement, and the motion is not simple harmonic.
The vibration of atoms in a solid — each atom is held in place by bonds that act like tiny springs.
A tuning fork — the prongs vibrate in SHM, producing a pure tone.
The Bottom Line
Simple Harmonic Motion is any motion driven by a restoring force that is proportional to and opposite the displacement. It produces a sinusoidal oscillation with a constant period that is independent of amplitude. Everything else — the equations, the graphs, the energy transformations — is just unpacking that single, elegant idea.
Looking up "Simple Harmonic Motion: Definition, Formula & Real-World Examples" or "Simple Harmonic Motion important questions 11" is a common way students land here, and rightly so — simple harmonic motion is a core part of the Class 11 Physics NCERT/CBSE curriculum. Expect it to reappear, often in a slightly disguised form, across JEE Main, NEET and state engineering/medical entrance exams.
Use the reference-circle (phasor) method: equal speed at A and B means the two phase angles are related by cosθ1=cosθ2; combining the two given 3 s intervals with this condition pins down ω.
✓Final answer
(c) 12 s
Step 1. Represent the SHM as x=Rsin(ωt+ε) on the reference circle, with phase θ=ωt+ε. Since SHM speed v=Rωcosθ depends only on cosθ, 'crossing A and B with the same velocity' after a direct transit A→B of 3 s means cosθ1=cos(θ1+3ω), which forces θ1+3ω=−θ1+2nπ for the non-trivial case, i.e. θ1=nπ−1.5ω.
Step 2. So the phase at B is θ2=θ1+3ω=nπ+1.5ω.
Step 3. 'Returns to B after another 3 s' means the particle is again at the same position B, but now moving the opposite way (it has turned around at the extreme in between), 3 s later. Same position with opposite-signed velocity corresponds to phase θ3=π−θ2(mod2π) (since sin(π−θ)=sinθ but cos(π−θ)=−cosθ). Since θ3=θ2+3ω, this gives 2θ2+3ω=π+2mπ.
Step 4. Substituting θ2 from Step 2: 2(nπ+1.5ω)+3ω=π+2mπ⇒6ω=π+2(m−n)π. Taking the smallest positive solution (m=n): 6ω=π⇒ω=π/6rad s−1.
Step 5. Time period T=2π/ω=2π/(π/6)=12 s.
✓Final answer
(c) 12 s
Use the reference-circle phasor picture: equal speed fixes a phase relation between the two crossing instants, and the 'returns after 3 s' condition fixes a second phase relation; solving both together gives omega and hence T.
Assuming A and B must be the mean position and an extreme, which is not given and not needed to solve the problem.
Treating 'returns to B after another 3 s' as simply another equal 3 s transit in the same direction, instead of a return passage with reversed velocity.