Waves — JEE Main Questions

1. A source of frequency 500 Hz moves at 20 m/s toward a stationary wall. An observer at the source hears the echo. Speed of sound is 340 m/s. What is the beat frequency heard?

1. 125 Hz
2. 31 Hz
3. 62 Hz
4. 94 Hz

Answer: 62 Hz

First, wall receives f1 = f v/(v - v_s) = 500 × 340/320 = 531.25 Hz. Then wall acts as stationary source (since wall is stationary) emitting f1. Observer moves with source at 20 m/s toward wall, so f2 = f1 (v + v_o)/v = 531.25 × 360/340 = 562.5 Hz. Beat = f2 - f = 62.5 Hz ≈ 62 Hz.

2. In a longitudinal sound wave, where is the pressure variation maximum?

1. At a displacement node
2. At a displacement antinode
3. Midway between node and antinode
4. At a point of zero slope

Answer: At a displacement node

In a longitudinal wave, pressure variation is proportional to the spatial derivative of displacement (∂P ∝ -∂s/∂x). At a displacement node, the slope of displacement is maximum, so pressure variation is maximum. Hence, displacement node corresponds to pressure antinode.

3. A tuning fork of frequency 256 Hz produces 4 beats per second with another fork. When the unknown fork is loaded with wax, the beat frequency becomes 6 Hz. What is the frequency of the unknown fork?

1. 252 Hz
2. 260 Hz
3. 262 Hz
4. 250 Hz

Answer: 260 Hz

Initially, f = 256 ± 4, so 260 Hz or 252 Hz. Loading with wax lowers frequency. If f were 252 Hz, lowering it would reduce the difference, so beats would decrease. Since beats increase to 6 Hz, f must be 260 Hz (lowering it increases difference).

4. A car moving at 30 m/s sounds a horn of frequency 400 Hz. The sound reflects off a truck moving at 20 m/s away from the car. What frequency does the car driver hear? (v = 340 m/s)

1. 412 Hz
2. 424 Hz
3. 477 Hz
4. 451 Hz

Answer: 424 Hz

The echo undergoes two Doppler shifts. First, the truck receives f₁ = f × (v - v_t)/(v - v_c) = 400 × 320/310 ≈ 412.9 Hz. Then the truck reflects as a source moving away, so the car hears f₂ = f₁ × (v + v_c)/(v + v_t) = 412.9 × 370/360 ≈ 424 Hz.

5. A sonometer wire of linear mass density 0.01 kg/m resonates with a tuning fork at a length of 0.4 m. When the fork is loaded with wax, the beat frequency is 4 Hz when the wire length is changed to 0.41 m. Find the original frequency of the fork.

1. 200 Hz
2. 196 Hz
3. 204 Hz
4. 210 Hz

Answer: 200 Hz

Using f = (1/2L)√(T/μ), initial fork frequency f = (1/0.8)√(T/0.01). After loading, fork frequency decreases by Δf, and wire length 0.41 m gives f_wire' = (1/0.82)√(T/0.01). Beat frequency = |f - Δf - f_wire'| = 4 Hz. Solving yields f = 200 Hz.

6. A source of sound of frequency 500 Hz moves at 30 m/s toward a stationary observer. Speed of sound is 340 m/s. What is the observed frequency?

1. 500 Hz
2. 459 Hz
3. 548 Hz
4. 544 Hz

Answer: 548 Hz

Use the Doppler formula for a moving source and stationary observer: f' = f v/(v - v_s). Here v = 340 m/s, v_s = 30 m/s, f = 500 Hz. So f' = 500 * 340/(340 - 30) = 500 * 340/310 ≈ 548 Hz. The frequency increases because the source approaches.

7. In a resonance tube experiment at 27°C, the first resonance length is 16.0 cm. The speed of sound at 0°C is 330 m/s. What is the first resonance length at 127°C? (Ignore end correction.)

1. 19.2 cm
2. 16.0 cm
3. 17.3 cm
4. 18.5 cm

Answer: 18.5 cm

Speed of sound v ∝ √T. At 27°C (300 K), v = 330 × √(300/273) ≈ 346 m/s. At 127°C (400 K), v' = 330 × √(400/273) ≈ 400 m/s. For a closed pipe, first resonance L = v/(4f). Since f is constant, L ∝ v. So L' = L × v'/v = 16.0 × 400/346 ≈ 18.5 cm.

8. Two sources emit sound waves of frequencies 440 Hz and 442 Hz with equal amplitude. At a point where the path difference is zero, how many times per second does the intensity become maximum?

Answer: 2.0

The beat frequency equals the difference in source frequencies: |442 - 440| = 2 Hz. At zero path difference, the phase difference is only due to frequency difference, so intensity maxima occur at the beat frequency, i.e., 2 times per second.

9. Which of the following is a mechanical wave?

1. Sound wave in water
2. Radio wave in air
3. Light wave in vacuum
4. X-ray in vacuum

Answer: Sound wave in water

A mechanical wave requires a material medium to propagate. Sound wave in water travels through the water medium, making it a mechanical wave. Light, radio, and X-rays are electromagnetic waves that can travel through vacuum.

10. Two waves overlap in a medium. Which statement about the principle of superposition is correct?

1. The waves permanently change each other's amplitude after overlap
2. The waves reflect off each other at the overlap region
3. The net displacement is the sum of individual displacements
4. The net displacement is the product of individual displacements

Answer: The net displacement is the sum of individual displacements

The principle of superposition states that when two or more waves overlap, the resultant displacement at any point is the algebraic sum of the displacements of the individual waves. The waves emerge unchanged after overlap.

11. A tuning fork of 256 Hz and an unknown fork produce 4 beats/s. Loading the unknown with wax increases beats to 6/s. What is the unknown frequency?

1. 262 Hz
2. 252 Hz
3. 250 Hz
4. 260 Hz

Answer: 260 Hz

Beat frequency = |256 - f| = 4, so f = 260 or 252. Loading wax decreases f. Beats increase to 6, so new |256 - f'| = 6. Since f' < f, the difference from 256 increases, meaning f was originally > 256. Hence f = 260 Hz.

12. A string fixed at one end has a heavy bead attached to the other end. The bead's mass is much larger than the string's mass. Which statement about the standing wave on the string is correct?

1. The bead end behaves as a node because the bead's inertia prevents motion.
2. The bead end behaves as an antinode because the bead reflects waves without phase change.
3. The bead end behaves as a node because the wave undergoes a phase change of π/2.
4. The bead end behaves as an antinode because the bead moves with the string.

Answer: The bead end behaves as a node because the bead's inertia prevents motion.

A heavy bead has large inertia, so it hardly moves when a wave arrives. This makes the displacement nearly zero, so the end acts as a node. The reflected wave undergoes a phase change of π, similar to a fixed end.