For the harmonic travelling wave $y = 2\cos 2\pi(10t - 0.0080x + 3.5)$ where $x$ and $y$ are in cm and $t$ is second. What is the phase difference between the oscillatory motion at two points separated by a distance of (a) 4 m (b) 0.5 m (c) $\dfrac{\lambda}{2}$ (d) $\dfrac{3\lambda}{4}$ (at a given instant of time) (e) What is the phase difference between the oscillation of a particle located at $x$ = 100cm, at $t = T$ s and $t$ = 5 s?

For the harmonic travelling wave y = 2cos 2π(10t - – Class 11 Physics

NCERT Exemplar · Q33

y = 2\cos 2\pi(10t - 0.0080x + 3.5)

(a) 4 m

(b) 0.5 m

(c)

\dfrac{\lambda}{2}

(d)

\dfrac{3\lambda}{4}

(at a given instant of time)

(e) What is the phase difference between the oscillation of a particle located at

x

= 100cm, at

t = T

s and

t

= 5 s?

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Concept understanding — Simple Harmonic Motion

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 = -k x

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 ( $a = \frac{d^2x}{dt^2}$ ), the condition $F = -kx$ becomes:

m \frac{d^2x}{dt^2} = -k x

This is a differential equation. Its solution — the position as a function of time — is:

x(t) = A \cos(\omega t + \phi)

Where:

$\omega = \sqrt{\frac{k}{m}}$ is the angular frequency (radians per second). It tells you how fast the oscillation is.
$\phi$ is the phase constant (determines where in the cycle you start measuring time). …

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For the harmonic wave $y=2\cos2\pi(10t-0.0080x+3.5)$ , the wave number is $k=0.016\pi\ \text{rad/cm}$ and the angular frequency is $\omega=20\pi\ \text{rad/s}$ . The spatial phase differences are (a) $6.4\pi$ rad, (b) $0.8\pi$ rad, (c) $\pi$ rad, (d) $\dfrac{3\pi}{2}$ rad, and the temporal phase difference in (e) is $98\pi$ rad, which is a whole number of cycles -- so the two instants are effectively in the same phase.

Reading off the wave parameters

The phase of the wave is $\phi(x,t)=2\pi(10t-0.0080x+3.5)$ . Comparing with $\phi=\omega t-kx+\text{const}$ :

\omega = 2\pi\times10 = 20\pi\ \text{rad/s} \qquad k = 2\pi\times0.0080 = 0.016\pi\ \text{rad/cm}

T = \frac{2\pi}{\omega} = \frac1{10} = 0.1\ \text{s} \qquad \lambda = \frac{2\pi}{k} = 125\ \text{cm}

Spatial phase difference (at a given instant)

For two points separated by $\Delta x$ at the same time, $\Delta\phi = k\,\Delta x$ .

(a) $\Delta x = 4\ \text{m} = 400\ \text{cm}$ :

\Delta\phi = 0.016\pi\times400 = 6.4\pi\ \text{rad}

(b) $\Delta x = 0.5\ \text{m} = 50\ \text{cm}$ :

\Delta\phi = 0.016\pi\times50 = 0.8\pi\ \text{rad}

(c) $\Delta x = \dfrac\lambda2$ :

\Delta\phi = k\cdot\frac\lambda2 = \frac{2\pi}{\lambda}\cdot\frac\lambda2 = \pi\ \text{rad}

(d) $\Delta x = \dfrac{3\lambda}4$ :

\Delta\phi = k\cdot\frac{3\lambda}4 = \frac{3\pi}2\ \text{rad}

Tip

These spatial phase differences don't need to be reduced modulo $2\pi$ unless asked for an "equivalent" phase within one cycle -- report the actual computed value.

Temporal phase difference (at a fixed position)

(e) At $x=100\ \text{cm}$ , comparing $t_1=T=0.1\ \text{s}$ and $t_2=5\ \text{s}$ : …

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Q.For the harmonic travelling wave y=2cos⁡2π(10t−0.0080x+3.5)y = 2\cos 2\pi(10t - 0.0080x + 3.5)y=2cos2π(10t−0.0080x+3.5) where xxx and yyy are in cm and ttt is second. What is the phase difference between the oscillatory motion at two points separated by a distance of

Reading off the wave parameters

Spatial phase difference (at a given instant)

Temporal phase difference (at a fixed position)

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Q.For the harmonic travelling wave $y = 2\cos 2\pi(10t - 0.0080x + 3.5)$ where $x$ and $y$ are in cm and $t$ is second. What is the phase difference between the oscillatory motion at two points separated by a distance of

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