All Exams Test series for 1 year @ ₹349 only
Question

For a travelling harmonic wave  $y(x, t) = 2.0 \cos 2\pi(10 t - 0.0080 x + 0.35)$, where $x$ and $y$ are in cm and $t$ in s. The phase difference between oscillatory motion of two points separated by a distance of $0.5 \text{ m}$ is :

This question was previously asked in
NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
$0.8\pi \text{ rad}$

Travelling Wave Phase Difference Analysis

The equation for the travelling harmonic wave is given as:

$y(x, t) = 2.0 \cos 2\pi(10 t - 0.0080 x + 0.35)$

In this equation, $x$ and $y$ are in centimeters (cm), and $t$ is in seconds (s).

Phase Difference Calculation Steps

The phase of the wave at any point $x$ and time $t$ is the argument of the cosine function:

$\Phi(x, t) = 2\pi(10 t - 0.0080 x + 0.35)$

The phase difference $\Delta \Phi$ between two points $x_1$ and $x_2$ at the same time $t$ is calculated as:

$\Delta \Phi = \Phi(x_2, t) - \Phi(x_1, t)$

Substituting the expression for phase:

$\Delta \Phi = \left[ 2\pi(10 t - 0.0080 x_2 + 0.35) \right] - \left[ 2\pi(10 t - 0.0080 x_1 + 0.35) \right]$

Simplify the expression by canceling common terms:

$\Delta \Phi = 2\pi [ (10 t - 0.0080 x_2 + 0.35) - (10 t - 0.0080 x_1 + 0.35) ]$

$\Delta \Phi = 2\pi [ -0.0080 x_2 + 0.0080 x_1 ]$

Factor out the common term $0.0080$:

$\Delta \Phi = 2\pi \times 0.0080 (x_1 - x_2)$

Distance Unit Conversion

The distance between the two points is given as $0.5 \text{ m}$. Since the variable $x$ in the wave equation is in centimeters, we must convert the distance to centimeters:

$\Delta x = x_1 - x_2 = 0.5 \text{ m} \times \frac{100 \text{ cm}}{1 \text{ m}} = 50 \text{ cm}$

Final Phase Difference Calculation

Now substitute the distance $\Delta x$ into the phase difference formula. We consider the magnitude of the phase difference:

$|\Delta \Phi| = |2\pi \times 0.0080 \times \Delta x|$

$|\Delta \Phi| = 2\pi \times 0.0080 \times 50$

Calculate the product:

$|\Delta \Phi| = 2\pi \times 0.4$

$|\Delta \Phi| = 0.8\pi \text{ rad}$

The phase difference between the two points separated by $0.5 \text{ m}$ is $0.8\pi$ radians.

Was this answer helpful?

Similar Questions

  1. The sum of kinetic energy and potential energy of a simple pendulum bob is $0.02 \text{ joule}$. The speed of the simple pendulum bob at equilibrium position is approximately :
    (Consider mass of the bob = $20 \text{ g}$)
  2. Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is $m\text{ kg}$ and the spring constant is $k\text{ Nm}^{-1}$. At a given instant, the extension of the spring is $x\text{ meter}$ and the speed of the particle is $v\text{ ms}^{-1}$. On the $x-v$ plane, if the graph of $v$ as a function of $x$ is a circle, then the correct option is :
  3. For sound waves, if the number of nodes for the $5^{\text{th}}$ harmonic of an open-ended pipe is $n$ and that for the $9^{\text{th}}$ harmonic of the same pipe with one of its ends closed is $m$, the ratio $\frac{n}{m}$ is :
  4. A cylindrical cork of uniform density floats in a liquid of density $\rho_1$. If the cork is depressed slightly and released, it oscillates harmonically with time period T. If the same cork floats in another liquid of density $\rho_2$, then the similar oscillation has time period 2T. The value of $\rho_2/\rho_1$ is :
  5. For a simple pendulum, having time period T, the variation of kinetic energy (K.E.) with time (t) is represented by :

Important Questions from Oscillations and Waves

  1. Two tuning forks $A$ and $B$ are sounded together giving rise to $8$ beats in $2\text{ s}$. When fork $A$ is loaded with wax, the beat frequency is reduced to $4$ beats in $2\text{ s}$. If the original frequency of tuning fork $B$ is $380\text{ Hz}$ then original frequency of tuning fork $A$ is _________ $\text{Hz}$.
  2. 5 beats/second are heard when a turning fork is sounded with a sonometer wire under tension, when the length of the sonometer wire is either 0.95 m or 1 m. The frequency of the fork will be
  3. A sinusoidal wave of wavelength 7.5 cm travels a distance of 1.2 cm along the x-direction in 0.3 sec. The crest P is at x=0 at t=0 sec and maximum displacement of the wave is 2 cm. Which equation correctly represents this wave ?
  4. A particle is subjected to two simple harmonic motions as : 
    $x_1 = \sqrt{7} \sin 5t \text{ cm}$ and $x_2 = 2\sqrt{7} \sin \left(5t + \frac{\pi}{3}\right) \text{ cm}$ 
    where x is displacement and t is time in seconds. The maximum acceleration of the particle is $x \times 10^{-2} \text{ ms}^{-2}$. The value of x is :

  5. In the resonance experiment, two air columns (closed at one end) of $100 \text{ cm}$ and $120 \text{ cm}$ long, give $15 \text{ beats per second}$ when each one is sounding in the respective fundamental modes. The velocity of sound in the air column is:
Need Expert Advice?
More Questions from NEET

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App