The document JEE Main Previous year questions (2016-20): Waves Notes | EduRev is a part of the JEE Course 35 Years JEE Main & Advanced Question Bank: Physics Class 11.

All you need of JEE at this link: JEE

**Q.1. A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears 2 beats/s. The oscillation frequency of each tuning fork is f _{0 }= 1400 Hz and the velocity of sound in air is 350m/s. The speed of each tuning fork is close to (2020)(1) 1/2 m/s(2) 1 m/s(3) 1/4 m/s**

Ans.

Given that, beats, n = 2

f

v = 350 m/s

Observer receives sound from two force, one approaching and one receding.

So apparent frequency from the two forks will be

and

v

We know that, beat is given by n = f

Since, v

1 = f

v

So, the speed of each tuning fork is 1/4 m/s

Q.2. A 1m long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP is 300 m/s, the frequency difference between the fundamental and second harmonic of this pipe is _________ Hz. (2020)

Ans.

Given that L = 1m, ρ

Velocity of a sound in the medium is given by

Since,

So,

Frequency difference between fundamental and second harmonic of open organ pipe is given by

=

Q.3. A transverse wave travels on a taut steel wire with a velocity of v when tension in it is 2.06 × 10

(1) 2.50 × 10

(2) 5.15 × 10

(3) 30.5 × 10

(4) 10.2 × 10

Ans.

Given that

Velocity in the string is given by

Since, m and L are constant v ∝ T

So,

Therefore,

(1) 5.8I

(2) 0.2I

(3) 3I

(4) I

Ans.

Solution.

General equation of wave is

x = A sin(2πvt± ϕ)

Where A is the amplitude of wave, n is frequency of the wave and f is the phase angle of the wave.

We have three waves with phase angles 0, π/4, -π/4. So,

x

x

x

Resultant wave when we superimposed these three waves are

The amplitude of the resultant wave is (√2 + 1)A.

We know that I ∝ A

So,

Q.5. A wire of length L and mass per unit length 6.0 x 10

(1) 2.1 m

(2) 1.1 m

(3) 8.1 m

(4) 5.1 m

Given that

Resonant frequency is given by

So,

490 = 420 + ν/2L

Velocity of wave in a wire is given by

So,**Q 6. A musician using an open flute of length 50 cm produces second harmonic sound waves. A person runs toward the musician from another end of a hall at a speed of 10 km/h. If the wave speed is 330 m/s, the frequency heard by the running person shall be close to (2019)(1) 666 Hz(2) 753 Hz(3) 500 Hz(4) 333 HzAns:** (1)

Velocity of observer, v

Frequency observed by observer

= 335.56 × 2 = 671.12 ≈ 666 Hz

(1) 8/17

Ans:

As we know that,

Dividing Eq. (2) by Eq. (1), we get

[speed of sound v = 340 m/s]

(1) 6

(2) 4

(3) 7

(4) 5

Ans:

For closed organ pipe, resonate frequency is

f

Therefore, number of overtones = n – 1 = 7 – 1= 6

(1) 5 N

(2) 75 N

(3) 10 N

(4) 12.5 N

Ans:

We have,

y = 0.03 sin (450 t – 9x)

General wave equation is,

y = a sin (ωt – kx)

So,

ω = 450

k = 9

We also know that

Ans:

(1) g/30

(2) g/5

(3) g/10

(4) g/20

Ans:

(1)

When the car has accelerated velocity is

(2)

Dividing Eq. (2) by Eq. (1), we have

Using binomial expansion, we have

(1) 10.0 cm

(2) 33.3 cm

(3) 16.6 cm

(4) 20.0 cm

Ans:

Wavelength of wave v = fλ

Therefore, distance between two consecutive nodes = λ/2

(1) 322 m/s

(2) 341 m/s

(3) 335 m/s

(4) 328 m/s

Ans:

Let

For

For

From Eq. (1) and Eq. (2), we get

Ans:

Since rod is clamped at centre. So centre it behaves as node & end it behave as antinode.

So,

**Q 15. A tuning fork vibrates with frequency 256 Hz and gives one beat per second with the third normal mode of vibration of an open pipe. What is the length of the pipe? (Speed of sound in air is 340 ms ^{-1}) (2018)(1) 180 cm(2) 190 cm(3) 220 cm(4) 200 cmAns:** (4)

Third normal mode of frequency in open pipe,

f = 3V

Where, V

Get L = 2m

Or

L = 200 cm

(1) 32cm

(2) 40cm

(3) 28cm

(4) 36cm

Answer:

Solution:

Given : e = 1 cm

For first resonance

For second resonance

⇒ ℓ

(1) 428 Hz

(2) 430 Hz

(3) 422 Hz

(4) 420 Hz

Ans:

Frequency of B is either 420Hz or 430Hz. As tension in B is increased its frequency will increase. If frequency is 430Hz, beat frequency will increase. If frequency is 420 Hz beat frequency will decrease, hence correct answer is 420Hz

(1) 17.3 GHz

(2) 15.3 GHz

(3) 10.1 GHz

(4) 12.1 GHz

Ans:

For relativistic motion

;

v = relative speed of approach

They are joined together at the point O, as shown in the figure. The combination is used as a sonometer wire and kept under tension T. The point O is midway between the two bridges. When a stationary wave is set up in the composite wire, the joint is found to be a node. The ratio of the number of antinodes formed in W

(1) 4 : 1

(2) 1 : 2

(3) 1 : 1

(4) 1 : 3

Ans:

What is the speed of the travelling wave moving in the positive x direction ? (x and t are in meter and second, respectively.) (2017)

(1) 120 m/s

(2) 90 m/s

(3) 160 m/s

(4) 180 m/s

Ans:

(take g = 10 ms

Ans

Solution:

(1) f/2

(2) 3f/4

(3) 2f

(4) f

Ans:

(1) 540 Hz

(2) 648 Hz

(3) 270 Hz

(4) 450 Hz

Ans:

(1) 340Hz

(2) 170 Hz

(3) 510 Hz

(4) 680 Hz

Ans:

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

46 docs|30 tests

- Test: MCQs (One or More Correct Option): Waves | JEE Advanced
- Test: Single Correct MCQs: Waves | JEE Advanced
- Test: 35 Year JEE Previous Year Questions: Waves
- Fill in the Blanks: Waves | JEE Advanced
- Subjective Type Questions: Waves | JEE Advanced
- Matrix-Match & Integer Answer Type Questions: Waves | JEE Advanced