Question 1

A sonometer wire of length ℓ has fundamental frequency f0. A bridge is placed at a point Δℓ away from the midpoint of the wire, where Δℓ is much smaller than ℓ. If the two resulting segments are made to vibrate in their fundamental modes, the number of beats produced is—

Accepted Answer

8f0Δℓ/ℓ. Segments l/2 + dl and l/2 - dl give beats f2 - f1 = (v/2)*(2*dl)/((l/2)^2) = 4*v*dl/l^2. With v = 2*f0*l this equals 8*f0*dl/l.

Question 2

Two identical piano wires are stretched with the same tension T. Each wire has a fundamental frequency of 600 Hz. If the tension of one wire is increased slightly, what fractional increase in tension is needed so that the two wires produce 6 beats per second when sounded together?

Accepted Answer

0.02. Since f ~ sqrt(T), f'/f = sqrt(T'/T). For 6 beats f' = 606 Hz, so T'/T = (606/600)^2 = 1.0201. The fractional increase in tension is about 0.02.

Question 3

Two sound sources, each producing waves of wavelength λ, are held at a fixed separation. An observer travels with speed u along the straight line joining the two sources. The beat frequency heard by the observer per second is

Accepted Answer

2u/λ. Moving toward one source gives f1 = (v+u)/lambda; moving away from the other gives f2 = (v-u)/lambda. Beat frequency = f1 - f2 = 2u/lambda.

Question 4

A body of specific gravity ρ is suspended from a light steel wire. The wire supports transverse standing waves with a fundamental frequency of 300 Hz. The body is then placed in water such that exactly half of its volume is below the surface. What is the new fundamental frequency of the wi

Accepted Answer

300((2ρ−1)/(2ρ))¹/2. f ~ sqrt(T). Initial tension = rho*rho_w*V*g; in water (half submerged) tension = rho_w*V*g*(rho - 1/2). So f_new = 300*sqrt((rho-1/2)/rho) = 300*((2rho-1)/(2rho))^(1/2).

Question 5

A transverse wave on a string has displacement y(x,t) = e^(-(ax² + bt² + 2abxt)). This corresponds to a:

Accepted Answer

wave travelling in the negative x-direction with speed √(b/a). The given displacement function represents a wave that has a specific form indicating it travels in the negative x-direction, as evidenced by the presence of the term involving both x and t in the exponent. The speed of the wave can be derived from the coefficients in the exponent, leading to the conclusion that it travels with speed √(b/a).

Question 6

A sound source A emits waves of frequency 1800 Hz and is moving downward toward the ground with terminal speed v. An observer B standing on the ground directly below the source detects the sound at 2150 Hz. The frequency of the waves reflected back from the ground and received by source A

Accepted Answer

2500 Hz. From f_obs = f*c/(c-v): 2150 = 1800*343/(343-v) gives v ~ 55.8 m/s. The ground receives 2150 Hz and re-emits it; source A (moving toward it) hears 2150*(c+v)/c = 2150*398.8/343 ~ 2500 Hz.

Question 7

In an FM signal, the frequency deviation is 18.75 kHz. If the transmission is carried in the 88–108 MHz band, what is the modulation percentage?

Accepted Answer

25%. The modulation percentage in frequency modulation is calculated by dividing the frequency deviation by the maximum frequency of the modulating signal. In this case, the maximum frequency is half the bandwidth of the FM signal (20 kHz for a 88-108 MHz band), leading to a modulation percentage of 18.75 kHz / 75 kHz = 0.25, or 25%.

Question 8

In an AM signal, the greatest peak-to-peak voltage observed is 24 mV and the least peak-to-peak voltage observed is 8 mV. What is the modulation index?

Accepted Answer

50%. The modulation index in AM signals is calculated as the ratio of the peak-to-peak voltage of the modulating signal to the peak-to-peak voltage of the carrier signal. In this case, the difference between the greatest and least peak-to-peak voltages indicates the modulation depth, leading to a modulation index of 50%.

Question 9

A broadcasting station operates on two bands: AM at 1020 kHz and FM at 89.5 MHz. For efficient reception, which antenna arrangement is appropriate?

Accepted Answer

Use a longer antenna for the AM band and a shorter one for the FM band. Antenna size scales with wavelength. AM at 1020 kHz has a much longer wavelength than FM at 89.5 MHz, so AM needs a longer antenna and FM a shorter one.

Question 10

For a 10 MHz radio wave to be reflected by the ionosphere in sky-wave communication, what is the least electron density required in the ionospheric layer?

Accepted Answer

About 1.2 × 10¹² m⁻³. The correct option is right because the minimum electron density required for reflecting a 10 MHz radio wave is approximately 1.2 × 10¹² m⁻³, which allows the wave to encounter sufficient free electrons to reflect back to the Earth's surface, enabling effective sky-wave communication.

Question 11

In a resonance-tube experiment used to determine the speed of sound at room temperature, the tube is first filled with water and then with glycerine, giving readings v1 and v2 respectively. Which statement is correct in this situation?

Accepted Answer

v1 and v2 are independent of the liquid used in the tube. The speed of sound in a resonance tube is determined by the properties of the gas above the liquid, not the liquid itself. Therefore, the readings v1 and v2, which correspond to different liquids, will not affect the speed of sound measured in the air above them.

Question 12

A tuning fork of frequency 256 Hz produces 5 beats per second with a vibrating piano string. When the tension in the piano string is increased slightly, the beat frequency falls to 2 beats per second. What was the frequency of the piano string before the tension was increased?

Accepted Answer

(256−5) Hz. The initial beat frequency of 5 Hz indicates that the piano string's frequency was 5 Hz lower than the tuning fork's frequency of 256 Hz, making it 251 Hz. When the tension is increased, the frequency of the string rises, resulting in a new beat frequency of 2 Hz, confirming that the original frequency was indeed 251 Hz.

Question 13

A particle in a medium has displacement given by y = 10⁻⁶ sin(100t + 20x + π/4) m, where t is measured in seconds and x in metres. What is the speed of the wave?

Accepted Answer

5 m/s. The speed of a wave can be determined from its wave equation, where the general form is y = A sin(ωt + kx). Here, ω (angular frequency) is 100 rad/s and k (wave number) is 20 rad/m. The wave speed v is given by v = ω/k, which calculates to 5 m/s.

Question 14

Two tuning forks are struck together and produce 4 beats each second. Then a small piece of tape is added to tuning fork 2, and when both forks are sounded again, the beat frequency becomes 6 per second. If tuning fork 1 has a frequency of 200 Hz, what was the initial frequency of tuning f

Accepted Answer

196 Hz. The initial beat frequency of 4 Hz indicates that the frequency of tuning fork 2 is either 196 Hz or 204 Hz, since it is close to 200 Hz. After adding tape, the beat frequency increases to 6 Hz, which means the frequency of tuning fork 2 must have decreased, confirming that the initial frequency was 196 Hz.

Question 15

An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency?

Accepted Answer

20%. For an observer moving toward a stationary source, f' = f(v + v0)/v = f(1 + 1/5) = 1.2f. The apparent frequency rises by 20%.

Question 16

A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with speed v m s⁻¹. The velocity of sound in air is 300 m s⁻¹. If the person can hear frequencies upto a maximum of 10,000 Hz, the maximum value of v upto which he can hear whistle is

Accepted Answer

15 m s⁻¹. The maximum frequency the person can hear is 10,000 Hz, and using the Doppler effect formula, we can determine the speed at which the whistle must be moving to produce this frequency when approaching the listener. By solving the equation, we find that the maximum speed v that allows the person to still hear the whistle is 15 m/s.

Question 17

A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency of this string is

Accepted Answer

105 Hz. Consecutive resonant frequencies of a fixed string differ by the fundamental: f0 = 420 - 315 = 105 Hz, which is the lowest resonant frequency.

Question 18

A sound absorber attenuates the sound level by 20 dB. The intensity decreases by a factor of

Accepted Answer

100. Level drop = 10*log10(I0/I). 20 = 10*log10(ratio) -> ratio = 10^2 = 100. The intensity decreases by a factor of 100.

Question 19

While measuring the speed of sound by performing a resonance column experiment, a student gets the first resonance condition at a column length of 18 cm during winter. Repeating the same experiment during summer, she measures the column length to be x cm for the second resonance. Then

Accepted Answer

x > 54. First resonance (winter): L = lambda/4 = 18 cm -> lambda = 72 cm, so 3*lambda/4 = 54 cm. In summer the sound speed and hence wavelength are larger, so the second-resonance length x > 54 cm.

Question 20

A wave travelling along the x-axis is described by the equation y(x,t) = 0.005 cos (αx − βt). If the wavelength and the time period of the wave are 0.08 m and 2.0 s, respectively, then α and β in appropriate units are

Accepted Answer

α = 25.00π, β = π. alpha = 2*pi/lambda = 2*pi/0.08 = 25*pi rad/m, and beta = 2*pi/T = 2*pi/2.0 = pi rad/s. So alpha = 25.00*pi, beta = pi.

Question 21

Three sound waves of equal amplitudes have frequencies (ν−1), ν, (ν+1). They superpose to give beats. The number of beats produced per second will be:

Accepted Answer

2. The pairwise beat frequencies are 1, 1, and 2 Hz; the maximum number of beats per second is between (v+1) and (v-1), i.e. 2 Hz. Answer: 2.

Question 22

A motor cycle starts from rest and accelerates along a straight path at 2 m/s². At the starting point of the motor cycle there is a stationary electric siren. How far has the motor cycle gone when the driver hears the frequency of the siren at 94% of its value when the motor cycle was at r

Accepted Answer

98 m. The motorcycle accelerates and approaches the stationary siren, causing a Doppler effect that shifts the frequency heard by the driver. When the frequency is perceived at 94% of its original value, the motorcycle's speed relative to the sound wave allows us to calculate the distance traveled using the relationship between speed, acceleration, and time.

Question 23

The equation of a wave on a string of linear mass density 0.04 kg m⁻¹ is given by y = 0.02(m) sin [2π ( t/(0.04(s)) − x/(0.50(m)))]. The tension in the string is

Accepted Answer

6.25 N. From y = 0.02 sin[2pi(t/0.04 - x/0.50)], wave speed v = lambda/T = 0.50/0.04 = 12.5 m/s. Tension T = mu*v^2 = 0.04 * 12.5^2 = 6.25 N.

Question 24

The transverse displacement y(x,t) of a wave on a string is given by y(x,t) = e^(−(ax² + bt² + 2√ab xt)). This represents a:

Accepted Answer

wave moving in −x direction with speed √(b/a). The given equation represents a wave that is influenced by both space and time variables, indicating it is a traveling wave. The negative sign in the exponent associated with the term involving xt suggests that the wave is propagating in the negative x direction, and the speed of this wave can be derived from the coefficients of the equation, resulting in a speed of √(b/a).

Question 25

A travelling wave represented by y = A sin (ωt − kx) is superimposed on another wave represented by y = A sin (ωt + kx). The resultant is

Accepted Answer

A standing wave having nodes at x = (n + 1/2) λ/2, n = 0,1,2..... sin(wt-kx)+sin(wt+kx) = 2A sin(wt) cos(kx), a standing wave. Nodes occur where cos(kx)=0, i.e. kx = (n+1/2)pi, so x = (n+1/2) lambda/2.

Question 26

Statement - 1: Two longitudinal waves given by equations: y1(x,t) = 2a sin(ωt − kx) and y2(x,t) = a sin (2ωt − 2kx) will have equal intensity.
Statement - 2: Intensity of waves of given frequency in same medium is proportional to square of amplitude only.

Accepted Answer

Statement-1 is true, statement-2 is false.. Statement-1 is true because the two waves have different amplitudes and frequencies, which affects their intensity. Statement-2 is false because intensity is not solely proportional to the square of amplitude when considering different frequencies; the frequency also plays a significant role in determining intensity.

Question 27

A cylindrical tube, open at both ends, has a fundamental frequency f in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air-column is now:

Accepted Answer

f. The fundamental frequency of a tube open at both ends is determined by the length of the air column. When half of the tube is submerged in water, the length of the air column remains unchanged, thus the fundamental frequency remains the same.

Question 28

A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.

Accepted Answer

6. The number of natural oscillations in a closed pipe can be determined using the formula for the fundamental frequency and its harmonics. For a closed pipe, the frequencies are given by the formula fₙ = n(v/4L), where n is an odd integer. Given the length of the pipe and the speed of sound, we can calculate the maximum harmonic that fits below 1250 Hz, which results in 6 possible oscillations.

Question 29

A train is moving on a straight track with a speed of 20 m s⁻¹. It is blowing its whistle at the frequency of 1000 Hz. The percentage change in frequency heard by a person standing near the track as the train passes him is (speed of sound = 320 m s⁻¹) close to:

Accepted Answer

12%. The percentage change in frequency is calculated using the Doppler effect formula, which accounts for the relative motion between the source of sound (the train) and the observer. As the train approaches, the frequency increases, and as it moves away, it decreases, leading to a net change that can be expressed as a percentage of the original frequency, resulting in a 12% change.

Question 30

A pipe open at both ends has a fundamental frequency f in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now:

Accepted Answer

f. When the pipe is half submerged in water, the effective length of the air column that vibrates is halved, but since the pipe is open at both ends, the fundamental frequency remains the same. The frequency is determined by the speed of sound in air and the length of the air column, and since the length does not change in terms of the fundamental frequency calculation, it remains f.

Question 31

A message signal of the form A cos(ωt) is sent by amplitude modulation using the carrier v0 sin(ω0 t). Which of the following represents the AM signal?

Accepted Answer

v0 sin(ω0 t) + (A/2) sin[(ω0 − ω)t] + (A/2) sin[(ω0 + ω)t]. AM signal = (v0 + A cos wt) sin w0 t = v0 sin w0 t + (A/2) sin[(w0+w)t] + (A/2) sin[(w0-w)t], giving carrier plus two sidebands.

Question 32

A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are:
(1) 2 MHz only
(2) 2005 kHz, and 1995 kHz
(3) 2005 kHz, 2000 kHz and 1995 kHz
(4) 2000 kHz, and 1995 kHz

Accepted Answer

2005 kHz, 2000 kHz and 1995 kHz. The correct option includes the carrier frequency (2 MHz or 2000 kHz) and the sidebands created by the modulation process, which are 5 kHz above and below the carrier frequency, resulting in frequencies of 2005 kHz and 1995 kHz.

Question 33

A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is: (take g = 10 m s⁻²)

Accepted Answer

2√2 s. The speed of a wave on a string is determined by the tension and mass per unit length. For a uniform string, the wave speed can be calculated using the formula v = √(T/μ), where T is the tension and μ is the mass per unit length. Given the length of the string and the acceleration due to gravity, the time taken for the wave pulse to travel the entire length of the string is derived as 2√2 s.

Question 34

A pipe open at both ends has a fundamental frequency in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now:

Accepted Answer

f. When the pipe is half submerged in water, the effective length of the air column is halved, but since the pipe is open at both ends, the fundamental frequency remains the same as it depends on the speed of sound in air and the length of the air column, which is unchanged in terms of its open-end configuration.

Question 35

The end correction of a resonance column is 1 cm. If the shortest length resonating with the tuning fork is 10cm, the next resonating length should be -

Accepted Answer

32 cm. With end correction e=1 cm, first resonance gives lambda/4 = L1+e = 10+1 = 11 cm, so lambda = 44 cm. Next resonance: 3*lambda/4 = L2+e = 33 cm -> L2 = 32 cm.

Question 36

Two sitar strings, A and B, playing the note 'Dha' are slightly out of tune and produce beats and frequency 5 Hz. The tension of the string B is slightly increased and the beat frequency is found to decrease by 3 Hz. If the frequency of A is 425 Hz, the original frequency of B is -

Accepted Answer

420 Hz. Beat = |fA - fB| = 5, so fB = 420 or 430 Hz. Increasing B's tension raises fB; the beat frequency decreased to 2 Hz, meaning fB moved closer to 425 Hz. That only happens if fB started below A, so the original frequency of B is 420 Hz.

Question 37

A plane polarized monochromatic EM wave is traveling in a vacuum along z direction such that at t = t1 it is found that the electric field is zero at a spatial point z1. The next zero that occurs in its neighbourhood is at z2. The frequency of the electromagnetic wave is

Accepted Answer

1.5 × 10⁸ / |z2 − z1|. Adjacent zeros of the field are separated by half a wavelength, so lambda = 2|z2 - z1|. The frequency f = c/lambda = (3x10^8)/(2|z2 - z1|) = 1.5x10^8 / |z2 - z1|.

Question 38

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 -

Accepted Answer

195 Hz. The frequency of the tuning fork can be determined using the relationship between the beats heard and the frequencies of the two sound sources. Since 5 beats per second indicate a difference of 5 Hz between the fork and the wire's fundamental frequency, and the wire's frequencies for the given lengths are close to 200 Hz, the correct frequency of the fork is 195 Hz.

Question 39

A tuning fork vibrates with frequency 256 Hz and gives one best 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 = 340 m s−1) [JEE-Main On line-2018]

Accepted Answer

200 cm. The length of the pipe can be calculated using the formula for the fundamental frequency of an open pipe, which is given by f = v / (2L). Rearranging this gives L = v / (2f). Substituting the speed of sound (340 m/s) and the frequency (256 Hz) results in a length of 200 cm, confirming that option D is correct.

Question 40

The number of amplitude modulated broadcast stations that can be accommodated in a 300 kHz band width with the highest modulating frequency 15 kHz will be: [JEE-Main On line-2018]

Accepted Answer

10. The number of amplitude modulated stations that can fit within a given bandwidth is determined by the formula: Number of stations = Bandwidth / (2 x Highest modulating frequency). In this case, with a bandwidth of 300 kHz and a highest modulating frequency of 15 kHz, the calculation yields 10 stations.

Question 41

A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be (Assume that the highest frequency a person can hear is 20,000 Hz)

Accepted Answer

6. A closed pipe supports odd harmonics: 1.5, 4.5, 7.5, 10.5, 13.5, 16.5, 19.5 kHz are all <= 20 kHz (7 frequencies). The fundamental is 1.5 kHz, so the number of overtones distinctly heard = 7 - 1 = 6.

Question 42

A heavy ball of mass M is suspended from the ceiling of a car by a light string of mass m (m << M). When the car is at rest, the speed of transverse waves in the string is 60 m s⁻¹. When the car has acceleration a, the wave-speed increases to 60.5 m s⁻¹. The value of a, in terms of gravita

Accepted Answer

g/5. At rest T=Mg, v=60. Accelerating, T=M*sqrt(g^2+a^2). Since v^2 ~ T, (60.5/60)^2 = sqrt(1+(a/g)^2) -> 1.01674 = sqrt(1+(a/g)^2). Squaring: (a/g)^2 = 0.0338, a/g = 0.184. Closest option is g/5.

Question 43

A musician using an open flute of length 50 cm produces second harmonic sound waves. A person runs towards 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:

Accepted Answer

666 Hz. Open flute 2nd harmonic: f = 2*v/(2L) = 2*330/(2*0.5) = 660 Hz. Observer moves toward source at 10 km/h = 2.78 m/s: f' = f*(v+vo)/v = 660*(330+2.78)/330 ~ 666 Hz.

Question 44

The pressure wave, P = 0.01 sin [1000t - 3x] N m⁻², corresponds to the sound produced by vibrating blade on a day when atmospheric temperature is 0°C. On some other day when temperature is T, the speed of sound produced by the same blade at the same frequency is found to be 336 m s⁻¹. A pr

Accepted Answer

4°C. At 0 C, v0 = 1000/3 = 333.3 m/s. Since v is proportional to sqrt(T), T = 273*(336/333.3)^2 ≈ 277.4 K ≈ 4 C, so the answer is 4 C.

Question 45

A string is clamped at both the ends and it is vibrating in its 4th harmonic. The equation of the stationary wave is y = 0.3 sin(0.157x) cos(200πt). The length of the string is: (All quantities are in SI units.)

Accepted Answer

(2) 80 m. In the 4th harmonic of a clamped string, there are 4 segments (or loops) along the length of the string. The wavelength can be determined from the wave equation, where the wave number k = 0.157 m⁻¹ gives a wavelength of 2π/k. Since the length of the string is equal to 4 times the wavelength divided by 4 (for the 4 segments), the total length calculates to 80 m.

Question 46

Two cars A and B are moving away from each other in opposite directions. Both the cars are moving with a speed of 20 m s⁻¹ with respect to the ground. If an observer in car A detects a frequency 2000 Hz of the sound coming from car B, what is the natural frequency of the sound source in ca

Accepted Answer

2250 Hz. With v=340, vo=vs=20: 2000 = f0*(340-20)/(340+20) -> f0 = 2000*360/320 = 2250 Hz.

Question 47

Equation of travelling wave on a stretched string of linear density 5 g/m is y = 0.03 sin(450 t - 9x) where distance and time are measured in SI units. The tension in the string is:

Accepted Answer

12.5 N. The tension in the string can be calculated using the wave speed and linear density. The wave speed is derived from the angular frequency and wave number in the equation, leading to a tension of 12.5 N when applying the formula T = μv², where μ is the linear density.

Question 48

A resonance tube is old and has jagged end. It is still used in the laboratory to determine velocity of sound in air. A tuning fork of frequency 512 Hz produces first resonance when the tube is filled with water to a mark 11 cm below the reference mark, near the open end of the tube. The e

Accepted Answer

328 m s⁻¹. With unknown end correction e: L1+e=v/(4*512), L2+e=v/(4*256). Subtracting, L2-L1=0.16 m = v/2048, giving v=0.16*2048 = 328 m/s.

Question 49

Two identical strings X and Z made of same material have tension Tₓ and T_z in them. Their fundamental frequencies are 450 Hz and 300 Hz, respectively, then the ratio Tₓ/T_z is:

Accepted Answer

2.25. For identical strings f ~ sqrt(T), so Tx/Tz = (fx/fz)^2 = (450/300)^2 = 1.5^2 = 2.25.

Question 50

For a transverse wave travelling along a straight line, the distance between two peaks (crests) is 5m, while the distance between one crest and one trough is 1.5 m. The possible wavelengths (in m) of the waves are

Accepted Answer

1/1, 1/3, 1/5,..... The correct option represents the wavelengths of the wave in terms of fractions, which are consistent with the relationship between the distance between crests and troughs. Since the distance between a crest and a trough is half the wavelength, the possible wavelengths must be in the form of fractions that reflect this relationship.

JEE Main Physics: Waves questions with solutions

Questions