Waves Previous Year Questions — JEE Main Physics
22 past-year Waves questions from JEE Main (Physics).
Q6 (2025)
A sinusoidal wave of wavelength $7.5 \ \mathrm{cm}$ travels a distance of $1.2 \ \mathrm{cm}$ along the x-direction in $0.3 \ \mathrm{sec}$. The crest P is at $x = 0$ at $t = 0 \ \mathrm{sec}$ and maximum displacement of the wave is $2 \ \mathrm{cm}$ . Which equation correctly represents this wave?
- $y = 2\cos (0.83x - 3.35t) \ \mathrm{cm}$
- $y = 2\sin (0.83x - 3.5t) \ \mathrm{cm}$
- $y = 2\cos (3.35x - 0.83t) \ \mathrm{cm}$
- $y = 2\cos (0.13x - 0.5t) \ \mathrm{cm}$
### Related Formula
1. Wave function (moving along $+x$ direction) with a peak at $x=0, t=0$:
$$y(x, t) = A \cos(kx - \omega t)$$
2. Wave number:
$$k = \frac{2\pi}{\lambda}$$
3. Wave speed:
$$v = \frac{\omega}{k}$$
### Core Logic
Given parameters:
- Wavelength $\lambda = 7.5 \ \mathrm{cm}$
- Distance travelled $\Delta x = 1.2 \ \mathrm{cm}$ in $\Delta t = 0.3 \ \mathrm{s}$
- Maximum displacement (amplitude) $A = 2 \ \mathrm{cm}$
Let's calculate the wave parameters:
1. **Wave number ($k$):**
$$k = \frac{2\pi}{7.5} = \frac{20\pi}{75} = \frac{4\pi}{15} \approx 0.838 \ \mathrm{rad/cm}$$
2. **Wave speed ($v$):**
$$v = \frac{\Delta x}{\Delta t} = \frac{1.2}{0.3} = 4 \ \mathrm{cm/s}$$
3. **Angular frequency ($\omega$):**
$$\omega = v \cdot k = 4 \times \frac{4\pi}{15} = \frac{16\pi}{15} \approx 3.35 \ \mathrm{rad/s}$$
Since the crest is at $x=0$ at $t=0$, $y(0,0) = 2 = A$. This boundary condition demands a cosine function.
### Step 1: Write wave equation
Substitute $A$, $k$, and $\omega$ into the standard form:
$$y(x, t) = 2 \cos(0.83x - 3.35t) \ \mathrm{cm}$$
This perfectly matches Option (1).
### Pattern Recognition
Sees: Wavelength and speed to determine travelling wave equation.
Trap: Choosing sine instead of cosine. Since the crest (maximum displacement) is at $x=0, t=0$, $y(0,0)$ must equal $A$, which is satisfied only by the cosine function.
Shortcut: Calculate $k = 2\pi / 7.5 \approx 0.83$. This immediately eliminates Options (3) and (4). Calculate speed $v = 4$, so $\omega = 4 \times 0.83 \approx 3.35$, which points directly to Option (1).
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Waves
Q2 (2025)
In the resonance experiment, two air columns (closed at one end) of $100\mathrm{~cm}$ and $120\mathrm{~cm}$ long, give 15 beats per second when each one is sounding in the respective fundamental modes. The velocity of sound in the air column is :
- $335\mathrm{~m/s}$
- $370\mathrm{~m/s}$
- $340\mathrm{~m/s}$
- $360\mathrm{~m/s}$
### Related Formula
For an air column closed at one end, the fundamental frequency $f$ is given by:
$$f = \frac{v}{4l}$$
where $v$ is the velocity of sound and $l$ is the length of the air column.
### Core Logic
Given parameters:
- $l_1 = 100\mathrm{~cm} = 1.0\mathrm{~m}$
- $l_2 = 120\mathrm{~cm} = 1.2\mathrm{~m}$
- Beats per second ($f_1 - f_2$) = 15
### Step 1: Write the equation for beat frequency
Since $l_1 < l_2$, the frequency $f_1 > f_2$. Hence:
$$\text{Beat frequency} = f_1 - f_2 = \frac{v}{4l_1} - \frac{v}{4l_2}$$
$$15 = \frac{v}{4} \left( \frac{1}{l_1} - \frac{1}{l_2} \right)$$
### Step 2: Solve for velocity of sound ($v$)
Substitute the lengths in meters:
$$15 = \frac{v}{4} \left( \frac{1}{1.0} - \frac{1}{1.2} \right)$$
$$15 = \frac{v}{4} \left( 1 - \frac{5}{6} \right)$$
$$15 = \frac{v}{4} \left( \frac{1}{6} \right)$$
$$15 = \frac{v}{24}$$
$$v = 15 \times 24 = 360\mathrm{~m/s}$$
### Pattern Recognition
Beat problems involving standing waves in organ pipes can be calculated faster by remembering that $f \propto \frac{1}{l}$. This allows setting up the proportion $v = 4 \cdot \Delta f \cdot \frac{l_1 l_2}{l_2 - l_1}$ directly as a short-cut.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Waves
Q1 (2025)
Two harmonic waves moving in the same direction superimpose to form a wave $x = a \cos (1.5t) \cos (50.5t)$ where $t$ is in seconds. Find the period with which they beat (close to nearest integer)
- $6 \mathrm{~s}$
- $4 \mathrm{~s}$
- $1 \mathrm{~s}$
- $2 \mathrm{~s}$
### Related Formula
The product of cosines can be transformed into a sum using the trigonometric identity:
$$\cos A \cos B = \frac{1}{2} [\cos(A + B) + \cos(A - B)]$$
The beat frequency $f_{\text{beat}}$ is given by:
$$f_{\text{beat}} = |f_1 - f_2| = \left| \frac{\omega_1 - \omega_2}{2\pi} \right|$$
The beat period $T_{\text{beat}}$ is:
$$T_{\text{beat}} = \frac{1}{f_{\text{beat}}}$$
### Core Logic
Rewrite the superposition equation:
$$x = a \cos(1.5t) \cos(50.5t)$$
Apply the identity with $A = 50.5t$ and $B = 1.5t$:
$$x = \frac{a}{2} [\cos(52t) + \cos(49t)]$$
Here, the two component frequencies are:
$$\omega_1 = 52 \mathrm{~rad/s} \implies f_1 = \frac{52}{2\pi}$$
$$\omega_2 = 49 \mathrm{~rad/s} \implies f_2 = \frac{49}{2\pi}$$
Calculate the beat frequency:
$$f_{\text{beat}} = f_1 - f_2 = \frac{52 - 49}{2\pi} = \frac{3}{2\pi} \mathrm{~Hz}$$
### Step 1: Calculate Beat Period
The time period of beats is:
$$T_{\text{beat}} = \frac{1}{f_{\text{beat}}} = \frac{2\pi}{3} \approx \frac{2 \times 3.14}{3} = 2.09 \mathrm{~s}$$
Rounding to the nearest integer gives $2 \mathrm{~s}$.
### Pattern Recognition
Sees: product of two cosines with significantly different coefficients $\omega_1$ and $\omega_2$.
Shortcut: The beat period is simply $2\pi$ divided by the difference between the two component frequencies, where the component frequencies are $(\omega_{\text{average}} \pm \omega_{\text{envelope}})$. The difference is $2 \times \omega_{\text{envelope}} = 2 \times 1.5 = 3 \mathrm{~rad/s}$. Thus, $T = 2\pi / 3 \approx 2 \mathrm{~s}$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Waves
Q317 (2025)
The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves, $y_{1}(x,t) = 4\sin(kx - \omega t)$ and $y_{2}(x,t) = 2\sin\left(kx - \omega t + \frac{2\pi}{3}\right)$, are:
(Take the angular frequency of initial waves same as $\omega$)
- $\left[6, \frac{2\pi}{3}\right]$
- $\left[6, \frac{\pi}{3}\right]$
- $\left[\sqrt{3}, \frac{\pi}{6}\right]$
- $\left[2\sqrt{3}, \frac{\pi}{6}\right]$
### Related Formula
$$A_{\text{res}} = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos\phi}$$
$$\tan\theta = \frac{A_2 \sin\phi}{A_1 + A_2 \cos\phi}$$
where,
$A_1, A_2$ = amplitudes of individual harmonic waves
$\phi$ = phase difference between the waves
$A_{\text{res}}$ = resultant amplitude
$\theta$ = resultant phase angle relative to the first wave
### Core Logic
Given parameters:
- $A_1 = 4$
- $A_2 = 2$
- Phase difference, $\phi = \frac{2\pi}{3} = 120^\circ$
{{SOL_IMG1}}
Calculate Resultant Amplitude:
$$A_{\text{res}} = \sqrt{4^2 + 2^2 + 2(4)(2) \cos 120^\circ}$$
$$A_{\text{res}} = \sqrt{16 + 4 + 16 \left(-0.5\right)} = \sqrt{20 - 8} = \sqrt{12} = 2\sqrt{3}$$
Calculate Resultant Phase (angle $\theta$):
$$\tan\theta = \frac{2 \sin 120^\circ}{4 + 2 \cos 120^\circ} = \frac{2 \left(\frac{\sqrt{3}}{2}\right)}{4 + 2 \left(-0.5\right)} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$$
$$\theta = \frac{\pi}{6}$$
### Step 1: Result Format
Represent the resultant amplitude and phase as a pair:
$$\left[2\sqrt{3}, \frac{\pi}{6}\right]$$
### Pattern Recognition
Sees: Vector-like addition of wave amplitudes.
Shortcut: Solve it like a vector addition problem. Vector $A_1$ along horizontal ($0$), and Vector $A_2$ at $120^\circ$. The magnitude is $\sqrt{4^2+2^2-2(4)(2)(0.5)} = 2\sqrt{3}$ and direction is $\pi/6$. ✓
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Waves
Q13 (2025)
Two strings with circular cross section and made of same material, are stretched to have same amount of tension. A transverse wave is then made to pass through both the strings. The velocity of the wave in the first string having the radius of cross section $R$ is $v_{1}$, and that in the other string having radius of cross section $R/2$ is $v_{2}$. Then $\frac{v_{2}}{v_{1}} =$
- $\sqrt{2}$
- $2$
- $8$
- $4$
### Related Formula
$$v = \sqrt{\frac{T}{\mu}}$$
$$\mu = \rho A = \rho \left(\pi R^2\right)$$
where,
$v$ = velocity of transverse wave on a string
$T$ = tension on string
$\mu$ = linear mass density
$\rho$ = density of material
$A$ = cross-sectional area
### Core Logic
Since both strings are made of the same material, their density $\rho$ is the same.
Also, they are stretched to have the same amount of tension $T$.
Substitute the formula for $\mu$ into the velocity equation:
$$v = \sqrt{\frac{T}{\rho \pi R^2}} = \frac{1}{R} \sqrt{\frac{T}{\rho \pi}}$$
Thus, the wave velocity is inversely proportional to the radius of the cross section of the string:
$$v \propto \frac{1}{R}$$
$$\frac{v_2}{v_1} = \frac{R_1}{R_2}$$
### Step 1: Ratio Computation
Given:
- $R_1 = R$
- $R_2 = R/2$
Substitute these values:
$$\frac{v_2}{v_1} = \frac{R}{R/2} = 2$$
### Pattern Recognition
Sees: Circular cross-section strings + transverse wave speed relation.
Shortcut: Wave velocity $v \propto \frac{1}{\sqrt{\mu}} \propto \frac{1}{R}$. If the radius is halved, the mass per unit length decreases by 4 times, which makes the speed increase by $\sqrt{4} = 2$ times. ✓
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Waves
