427 + Interview Questions in WAVES IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Sound waves of frequency `660 Hz` fal normally on a perfectly reflecting wall. The shortest distance from the wall at which the pair particles have maximum amplitude of vibration is ……………… meters.
Answer» Correct Answer - A::B `c = vlambda` : .lambda = ( c )/(v) = (330)/(660) = 0.5m` the rarefaction will be at a distance of `(lambda)/(4) = (0.5)/(4) = 0.125m`

Discussion

2.	Calculate the frequency of `2^(nd)` harmonic in an open organ pipe of length 34 cm if the velocity of sound is 340m/s.
Answer» In an open organ pipe, the frequency of the second harmonic is `v_(2)=2xx(v)/(2L)=(2xx340)/(2xx0.34)=1000Hz`

Discussion

3.	What is the beat frequency, when two waves of frequency 450 Hz and 456 Hz are superposed?
Answer» Beat frequency `(v)=\|v_(1)-v_(2)\|` `=456Hz-450Hz` =6Hz

Discussion

4.	Velocity of sound in a medium is 490m/s. two waves of wavelength 98 cm and 100 cm are superposed. Calculate the beat frequency.
Answer» Correct Answer - 10

Discussion

5.	Two sound sources emitting sound each of wavelength `lambda` are fixed at a given distance apart. A listener moves with a velocity `u` along the line joining the two suorces. The number of beats heard by him per second isA. `(2u)/lambda`B. `u/lambda`C. `u/(2lambda)`D. `lambda/u`
Answer» Correct Answer - a

Discussion

6.	Two waves of wavelength 50 cm and 51 cm produce 12 beat/s . The speed of sound isA. 306 m/sB. 331 m/sC. 340 m/sD. 360 m/s
Answer» Correct Answer - a

Discussion

7.	A standing wave having 3 nodes and 2 antinodes is formed between two atoms having a distance 1.21Å between them. The wavelength of the standing wave isA. 1.21 ÅB. 1.42ÅC. 6.05ÅD. 3.63Å
Answer» Correct Answer - a

Discussion

8.	A vehicle, with a horn of frequency n is moving with a velocity of 30 m/s in a direction perpendicular to the straight line joining the observer and the vehicle. The observer perceives the sound to have a frequency `n + n_(1)`. Then (if the sound velocity in air is 300 m/s)A. `n_(1) = 10 n`B. `n_(1)=0`C. `n_(1)=0.1 n`D. `n_(1)=-0.1 n`
Answer» Correct Answer - b

Discussion

9.	A vehicle with a horn of frequency n is moving with a velocity of 30m/s in a direction perpendicular to the straight line joining the observer and the vehicle. The observer perceives the sound to have a frequency n + n1. Then (if the sound velocity in air is 300 m/s)(a) n1 = 10 n (b) n1 = 0 (c) n1 = – 0.1 n (d) n1 = 0.1 n
Answer» (b) n₁ = 0 No doppler effect is observed if the source moves perpendicular to the line joining the source and the observer. Therefore, the correct choice is (b).

9.

A vehicle with a horn of frequency n is moving with a velocity of 30m/s in a direction perpendicular to the straight line joining the observer and the vehicle. The observer perceives the sound to have a frequency n + n1. Then (if the sound velocity in air is 300 m/s)(a) n1 = 10 n (b) n1 = 0 (c) n1 = – 0.1 n (d) n1 = 0.1 n

Answer»

(b) n₁ = 0

No doppler effect is observed if the source moves perpendicular to the line joining the source and the observer. Therefore, the correct choice is (b).

Discussion

10.	Two organ pipes of lengths 65cm and70 cm respectively , are sounded simultaneosly. How many beats per second will be produced between the fundamental of the two pipes ? ( Velocity of sound ` = 330 m //s ) `
Answer» `l_(1) =65cm= 0.65m` `l_(2) = 70 cm = 0.7 m ` `v = 330 m //s ` No. of beats per second `Delta v = v_(1) - v_(2)` `= ( v)/(2l_(1))- (v)/(2l_(2)) = ( 330 )/( 2 xx 0.65) - ( 330)/( 2 xx 0.7)` `:. Delta v = 253.8 - 235 .8 = 18 Hz`

Discussion

11.	From a wave equation `y= 0.5 sin ((2pi)/3.2)(64t-x).` the frequency of the wave isA. 5 HzB. 15 HzC. 20 HzD. 25 Hz
Answer» Correct Answer - c

Discussion

12.	Two waves are approaching each other with a velocity of `20m//s` and frequency `n`. The distance between two consecutive nudes isA. `20/n`B. `10 / n`C. `5/n`D. `n/10`
Answer» Correct Answer - b

Discussion

13.	If `v_(1) , v_(2) and v_(3)` are the fundamental frequencies of three segments of stretched string , then the fundamental frequency of the overall string isA. `sqrtv=sqrtv_(1)+sqrtv_(2)+ sqrtv_(3)`B. `v=v_(1)+v_(2)+ v_(3)`C. `1/v=1/v_(1)+1/v_(2)+ 1/v_(3)`D. `1/sqrtv=1/sqrtv_(1)+1/sqrtv_(2)+ 1/sqrtv_(3)`
Answer» Correct Answer - c

Discussion

14.	A second harmonic has to be generated in a string of length l stretched between two rigid supports. The point where the string has to be plucked and touched areA. Plucked at `L/4` and touch at `L/2`B. Plucked at `L/4` and touch at `(3L)/2`C. Plucked at `L/2` and touch at `L/2`D. Plucked at `L/2` and touch at `(3L)/4`
Answer» Correct Answer - A The plucking point of the string will be the antinode and touching point will be node.

Discussion

15.	The wavelength of the note emitted by a tuning fork of frequency 100 Hz is 3.32 m. if the density of air at STP is 1.29 `kg//m^(3)`, the `gamma` for air isA. 1.57B. 2.3C. 1.41D. 1.732
Answer» Correct Answer - C

Discussion

16.	The change in phase if a wave is reflected from a rigid surface isA. zeroB. `pi` radianC. `2pi` radianD. `(pi)/(2)` radian
Answer» Correct Answer - B

Discussion

17.	The change in phase, if a wave is reflected at a less dense surface, isA. zeroB. `pi` radianC. `(pi)/(2)` radianD. `(2pi)/(3)` radian
Answer» Correct Answer - A

Discussion

18.	What is the phase difference between the incident and reflected wave when the wave is reflected by a rigid boundary.
Answer» `pi` Radian or `180^(@)`.

Discussion

19.	In case of a travelling wave, the reflection at a rigid boundary will take place with a phase change ofA. `pi/2` radianB. `pi/4` radianC. `pi/6` radianD. `pi` radian
Answer» Correct Answer - D In case of a travelling wave. the reflection at a rigid boundary will take place with a phase reversal or with a phase change of `pi` or `180^(@)`.

Discussion

20.	Which of the following statements are true for a stationary wave ?A. Every paticle has a fixed amplitude which is different from the amplitude of its nearest particleB. All the particles cross their mean position at the same timeC. There are some particles which are alwayss at restD. There is not transfer of energy across any plane
Answer» Correct Answer - A::B::D Consider the equation of a stationary wave ` y= a sin (kx) cos omegat` (a) clearly every paticle at x will have amplitude = a sin k x = fixed (b) for mean position y=0 `rArr " " cos omegat=0` `rArr " " omegat =(2n-1)(pi)/(2)` Hence for a fixed value of n, all partice are having same value of time `t=(2n-1)(pi)/(2omega)" "[because omega=` constant] (c) amplitude of all the particles are a sin (kx) which isi different particles at different values of x (d) the energy is a stationary wave is confined between two nodes (e) particles at different nodes are always at rest.

Discussion

21.	Four harmonic waves of equal freuencied and equal intensity `I_(0)` have phase angles `0, (pi)/(3), (2pi)/(3)` and `pi`. When they are superposed, the intensity of the resulting wave is `nI_(0)`. The value of `n` is
Answer» Correct Answer - C (3) `y = sqrt(I_(0))[sinO + sin((pi)/(3)) + sin((2pi)/(3)) + sinpi)]` `y = sqrt(I_(0))[sqrt(3)/(2) + sqrt(3)/(2)] = sqrt(3)sqrt(I_(0))` :. `I_(r) =y^(2) = 3I_(0)` rArr `n = 3`

Discussion

22.	A steel wire has a length of 12.0m and a mass of 2.10kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at `20^(@)C=343ms^(-1)`.
Answer» Length of the steel wire, l = 12 m Mass of the steel wire, m = 2.10 kg Velocity of the transverse wave, v = 343 m/s Mass per unit length, `mu=(m)/(l)=(2.10)/(12)=0.175 kg m^(-1)` For tension T, velocity of the transverse wave can be obtained using the relation : `v=sqrt((T)/(mu))` `therefore T=v^(2)mu` `=(343)^(2)xx0.175 = 20588~~2.06xx10^(4)N`

Discussion

23.	The displacement `y` of a wave travelling in the x-direction is given by `y = 10^(04) sin ((600t - 2x + (pi)/(3))meters` where `x` is expressed in meters and `t` in seconds. The speed of the wave-motion, in `ms^(-1)`, isA. (a) `300`B. (b) `600`C. ( c ) `1200`D. (d) `200`
Answer» Correct Answer - A (a) `y = 10^(-4) sin (600t - 2x + (pi)/(3))` But `y = A sin (omegat - kx + Phi)` On comparing we get `omega = 600`, `k = 2` `v = (omega)/(k) = (600)/(2) = 300 ms^(-1)`

Discussion

24.	The equation of a simple harmonic wave is given by `Y = 5sin""pi/2(100t - x)`, where x and y are in metre and time is in second. The time period of the wave (m seconds) will beA. 0.04B. 0.01C. 1D. 5
Answer» Correct Answer - A `y=5sin""pi/2(100t-x)` `y=Asin(omegat-kx)` `thereforeomega=(2pi)/T=pi/2xx100orT=2/50=0.04s`

Discussion

25.	A standing wave is represented by y = A sin (100t) cos (0.01x) where y and A are in millimetres, t in seconds and x in metres. The velocity of the wave is ………(a) 104 m/s (b) 1 m/s (c) 10-4 m/s (d) not derivable from the above information
Answer» (a) 10⁴ m/s v = \(\frac{ω}{k}\) = \(\frac{100}{0.10}\) = 10⁴ m/s

25.

A standing wave is represented by y = A sin (100t) cos (0.01x) where y and A are in millimetres, t in seconds and x in metres. The velocity of the wave is ………(a) 104 m/s (b) 1 m/s (c) 10-4 m/s (d) not derivable from the above information

Answer»

(a) 10⁴ m/s

v = \(\frac{ω}{k}\) = \(\frac{100}{0.10}\) = 10⁴ m/s

Discussion

26.	A standing wave is represeneted by `y=asin (100t)cos (0.01)x` in second and x is in ,metre. Velocity of wave isA. `10 ^(4)` m/sB. 1 m/sC. `10 ^(m/s`D. none of these
Answer» Correct Answer - a

Discussion

27.	Two tuning forks of frequencies 256 and 258 vibration /sec are sounded together , and then time interval between consecutive maxima heard by the observer is A ) 2 secB ) 0.5 secC ) 250 secD ) 252 sec
Answer» B ) 0.5 sec Explanations: n₁ = 256 n₂ = 258 Both tuning forks are making beats t_{( for max . )} = 1 / n₂ - n₁ sec = 1 / ( 258 - 256 ) = 1 / 2 = 0.5 sec

27.

Two tuning forks of frequencies 256 and 258 vibration /sec are sounded together , and then time interval between consecutive maxima heard by the observer is A ) 2 secB ) 0.5 secC ) 250 secD ) 252 sec

Answer»

B ) 0.5 sec

Explanations:

n₁ = 256

n₂ = 258

Both tuning forks are making beats

t_{( for max . )} = 1 / n₂ - n₁ sec

= 1 / ( 258 - 256 )

= 1 / 2

= 0.5 sec

Discussion

28.	Which of the following characteristics of sound help us in identifying two persons talking in a room without seeing them?A. LoudnessB. PitchC. QualityD. Intensity
Answer» Correct Answer - C

Discussion

29.	When a string fixed at its both ends vibrates in 1 loop, 2 loops, 3 loops and 4 loops, the frequencies are in the ratioA. 1 : 1 : 1 : 1B. 1 : 2 : 3 : 4C. 4 : 3 : 2 : 1D. 1 : 4 : 9 : 16
Answer» Correct Answer - B When the string vibrates in loops n, its frequency is, `upsilon_(n)=(nv)/(2L)` where L is the length of the string and v is the velocity of the wave `therefore` When the string fixed at its both ends vibrates in 1 loop. 2 loops, 3 loops and 4 loops, the frequencies are in the ratio 1:2:3:4.

Discussion

30.	A tuning fork vibrating with a frequency of 512Hz is kept close to the open end of a tube filled with water (Fig. 15.4). The water level in the tube is gradually lowered. When the water level is 17cm below the open end, maximum intensity of sound is heard. If the room temperature is 20°C, calculate(a) speed of sound in air at room temperature(b) speed of sound in air at 0°C(c) if the water in the tube is replaced with mercury, will there be any difference in your observations?
Answer» (a) 348.16 ms^-1 (b) 336 m/s (c) Resonance will be observed at 17cm length of air column, only intensity of sound heard may be greater due to more complete reflection of the sound waves at the mercury surface.

30.

A tuning fork vibrating with a frequency of 512Hz is kept close to the open end of a tube filled with water (Fig. 15.4). The water level in the tube is gradually lowered. When the water level is 17cm below the open end, maximum intensity of sound is heard. If the room temperature is 20°C, calculate(a) speed of sound in air at room temperature(b) speed of sound in air at 0°C(c) if the water in the tube is replaced with mercury, will there be any difference in your observations?

Answer»

(a) 348.16 ms^-1

(b) 336 m/s

(c) Resonance will be observed at 17cm length of air column, only intensity of sound heard may be greater due to more complete reflection of the sound waves at the mercury surface.

Discussion

31.	Show that when a string fixed at its two ends vibrates in 1 loop, 2 loops, 3 loops and 4 loops, the frequencies are in the ratio 1:2:3:4.
Answer» From the relation,v=nv/2L, the result follows.

Discussion

32.	A tuning fork vibrating with a frequency of 512Hz is kept close to the open end of a tube filled with water (Fig. 15.4). The water level in the tube is gradually lowered. When the water level is 17cm below the open end, maximum intensity of sound is heard. If the room temperature is 20°C, calculate(a) speed of sound in air at room temperature(b) speed of sound in air at 0°C(c) if the water in the tube is replaced with mercury, will there be any difference in your observations?
Answer» (a) 348.16 ms^-1 ^{(b) 336 m/s} (c) Resonance will be observed at 17cm length of air column, only intensity of sound heard may be greater due to more complete reflection of the sound waves at the mercury surface.

32.

A tuning fork vibrating with a frequency of 512Hz is kept close to the open end of a tube filled with water (Fig. 15.4). The water level in the tube is gradually lowered. When the water level is 17cm below the open end, maximum intensity of sound is heard. If the room temperature is 20°C, calculate(a) speed of sound in air at room temperature(b) speed of sound in air at 0°C(c) if the water in the tube is replaced with mercury, will there be any difference in your observations?

Answer»

(a) 348.16 ms^-1

^{(b) 336 m/s}

(c) Resonance will be observed at 17cm length of air column, only intensity of sound heard may be greater due to more complete reflection of the sound waves at the mercury surface.

Discussion

33.	A tuning fork vibrating with a frequency of 500 Hz is kept close to the open end of a tube filled with water, as shown in figure. The water level in the tube is gradually lowered When the water level is 17 cm below the open the open end, maximum intensity of sound is heard. If the room temperature is `20^(@)` C, the speed of sound in air at the temperature is A. `330ms^(-1)`B. `340ms^(-1)`C. `350ms^(-1)`D. `360ms^(-1)`
Answer» Correct Answer - B Here, `upsilon=500 Hz,L=17 cm=17xx10^(-2)m` as `upsilon=v/(4L)` `thereforev=upsilon(4L)=(500s^(-1))(4xx17xx10^(-2)m)=340ms^(-1)`

Discussion

34.	A tuning fork of frequency 440 Hz resonates with a tube closed at one end of length 1.8 cm and diameter 5 cm in fundamental mode. The velocity of sound in air isA. `336ms^(-1)`B. `343ms^(-1)`C. `300ms^(-1)`D. `350ms^(-1)`
Answer» Correct Answer - B In the fundamental mode of vibration, `(l+0.3D)=lamda/4=v/(4v)` `thereforev=4v(l+0.3D)=4xx440(0.18+0.3xx0.05)` `v=343ms^(-1)`

Discussion

35.	An organ pipe of cross-sectional area 100 `cm^(2)` resonates with a tuning fork of frequency 1000 Hz in fundamental tone. The minimum volume of water to be drained so the pipe again resonates with the same tuning fork is (Take velocity of wave = 320 m `s^(-1)`)A. `800 cm^(3)`B. `1200 cm^(3)`C. `1600 cm^(3)`D. `2000 cm^(3)`
Answer» Correct Answer - C Here, `upsilon=1000" Hz"` `1000=(v)/(4l_(1))=(3v)/(4l_(2))` Using `v= 320" m "s^(-1)`, we get, `l_(1)`= 8 cm and `l_(2)` = 24 cm `therefore" "` Minimum volume `=16xx100=1600" cm"^(3)`.

Discussion

36.	Two identical sinusoidal waves each of amplitude 10 mm with a phase difference of `90^(@)` are travelling in the same direction in a string. The amplitude of the resultant wave isA. 5 mmB. `10sqrt2` mmC. 15 mmD. 20 mm
Answer» Correct Answer - B Since the two waves are identical, they have amplitudes with phase difference of `90^(@)`. The amplitude of the resultant wave is, `A=sqrt(a^(2)+b^(2))=sqrt(a^(2)+a^(2)=asqrt2` Here, a= 10 mm, A=`10sqrt2`mm

Discussion

37.	Two tuning forks X and Y sounded together produce 10 beats per second. When Y is slightly loaded with wax, the number of beats reduces to 6 per second. If the frequency of X is 480 Hz, find that of Y.
Answer» Frequency of x = 480 Hz No. of beats with y = 10 ∴ Frequency of y = 480 +10 or 480-10 i.e., 490 or 470 On loading y, number of beats = 6 Frequency of y after loading = 486 or 474 ∴ The frequency of y before loading can not be 470, because if it were 470 before loading, it must be less than 470 after loading. ∴ Actual frequency of y = 490 Hz.

37.

Two tuning forks X and Y sounded together produce 10 beats per second. When Y is slightly loaded with wax, the number of beats reduces to 6 per second. If the frequency of X is 480 Hz, find that of Y.

Answer»

Frequency of x = 480 Hz

No. of beats with y = 10

∴ Frequency of y = 480 +10 or 480-10

i.e., 490 or 470

On loading y, number of beats = 6

Frequency of y after loading = 486 or 474

∴ The frequency of y before loading can not be 470, because if it were 470 before loading, it must be less than 470 after loading.

∴ Actual frequency of y = 490 Hz.

Discussion

38.	A steel wire has a length of 12 m and mass of 23.10 kg . What will be the speed of a transverse wave on this when a tension of `2.06xx10^(4)`N is applied?
Answer» Given, length of the wire `" " l=12 m` Mass of wire `" " m=2.10 kg` Tenstsion `" " T=2.06xx10^(4)N` Speed pf transverse wave `" " v=sqrt((T)/(mu))" "` [ where `mu`=mass per unit length ] `" " =sqrt((2.06xx10^(4))/(((2.10)/(12))))=sqrt((2.06xx12xx10^(4))/(2.10))=343m//s`

Discussion

39.	Out of the following gases under similar condition of temperature and pressure, the velocity of sound will be maximum inA. hydrogenB. nitrogenC. oxygenD. chlorine
Answer» Correct Answer - A Velocity of the sound in gas `sqrt((gammaRT)/M)` where the symbols have their usual meanings. All the given gases are diatomic and are at same temperature, `thereforevprop1/(sqrtM)`

Discussion

40.	The equation of a progressive wave can be given by Y = 15 sin (`660pit- 0.02pix`) cm. The frequency of the wave isA. 330 HzB. 342 HzC. 365 HzD. 660 Hz
Answer» Correct Answer - A `y=15sin(660pit-0.02pix)` The general equation of progressive wave is `y(x,t)=Asin((2pi)/Tt-(2pi)lamdax)` `therefore(2pi)/T=660piorI/T=330s^(-1)` or v=330Hz

Discussion

41.	A closes organ pipe of length 20 cm is sounded with a tuning fork in resonance. What is the frequency of the tuning fork? (v = 332 m/s)(a) 300 Hz (b) 350 Hz(c) 375 Hz (d) 415 Hz
Answer» (d) 415 Hz So, In resonance, the frequency of the fork is equal to the frequency of the organ pipe, \(f = \frac{v}{4L}\) = \(\frac{332}{4 \times 0.2}\) = 415 Hz

41.

A closes organ pipe of length 20 cm is sounded with a tuning fork in resonance. What is the frequency of the tuning fork? (v = 332 m/s)(a) 300 Hz (b) 350 Hz(c) 375 Hz (d) 415 Hz

Answer»

(d) 415 Hz

So,

In resonance, the frequency of the fork is equal to the frequency of the organ pipe,

\(f = \frac{v}{4L}\) = \(\frac{332}{4 \times 0.2}\) = 415 Hz

Discussion

42.	An air columbn in pipe, which is closed at one end, will be in resonance with a vibrating tuning fork of frequency `264 Hz` if the length of the column in `cm` is :A. (a) `31.25`B. ( b) `62.50`C. ( c ) `93.75`D. (d) `125`
Answer» Correct Answer - A::C (a,c) The wavelength possible in an air column in a pipe which has one closed end is `lambda = (4l)/((2n + 1))` So, `c = vlambda` `300 = 264 xx (4l)/(2n + 1)` `v = 264 Hz` as it is in resonnance with a vibrating tuning fork of frequency `264 Hz`. `l = (330 xx (2n +1))/(264 xx 4)` For `n = 1`, `l = 0.3125m = 31.25cm` For `n = 2`, `l = 0.9375m = 93.75cm`

Discussion

43.	Speed of sound waves in airA. is independet of temperatureB. increases with pressureC. increases with increases in humidityD. decreases with increase in humidity
Answer» Correct Answer - C Due to presence of moisture density of air decreases. We know that speed of sound in air is given by `v=sqrt((gammap)/(rho))` For any air `gamma and p` are constants. `" " vprop(1)/(sqrt(rho))` where `rho` is density of air. `(v_(2))/(v_(1))=sqrt((rho_(2))/(rho_(1))` where `rho_(1)` is density of dry air and `rho_(2)` is density of moist air. As `" " rho_(2)ltrho_(1)=(v_(2))/(v_(1))gt1rArrv_(2)gtv_(1)` Hence, speed of sound wave in air increases wiht in humidity.

Discussion

44.	Charge in temeprature of the medium changesA. frequency of sound wavesB. amplitude of sound wavesC. wavelength of sound wavesD. loudness of sound waves
Answer» Correct Answer - C Speed of sound wave in a medium `vpropsqrt(T)` (where T is temperature of the medium) Clearly, when temerature changes speed also changes. AS,` " " v=vlambda` where v is frequency and `lambda` is wavelength . Frequency (v) remains fixed `implies" " V=proplambda " or " lambdapropV` As does not change, so wavelength `(lambda)` changes.

Discussion

45.	A pipe 20cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source ? Will the same source be in resonance with the pipe if both ends are open ? ( speed of sound in air is `340ms^(-1)`)
Answer» Here, `l= 20 cm=0.2m , v_(n) = 430 Hz,v= 340 m//s ` The frequency of `n^(th) ` normal mode of vibration of closed pipe s `v_(n) =( 2n-1) (v)/( 4l)` `:. 430 = ( 2n-1) ( 340)/( 4 xx 0.2)` `2n-1= ( 430 xx 4xx 0.2)/( 340)= 1.02 ` `2n =2.02 , n = 1.01 ` Hence it will be the`1^(st)` normal mode of vibration . In a pipe, open at both ends we have `v_(n) =n xx( v)/( 2l) = ( n xx 340 )/( 2 xx 0.2) = 430` ` :. n = ( 430 xx 2 xx 0.2)/( 340 ) =0.5 ` As n has to be an integer, therefore open organ pipe cannot be in resonance with the source.

Discussion

46.	A pipe 20 cm long Is closed at one end. Which harmonic mode of the pipe 1s resonantly excited by a 430 Hz source? Will the same source be In resonance with the pipe if both ends are open? (speed of sound In air Is 340 m s-1 ).
Answer» Given : l = 20 cm = 0.2 m v = 340 ms^-1 f = 430 Hz For a pipe closed at one end : f = \(\frac{(2n-1)v}{4l}\) n = 1, 2, 3 …… ⇒ 430 = (2n-1)\(\frac{340}{4 \times 0.2}\) ⇒ n = 1 ⇒ Resonance occurs only for first/fundamental mode of vibration. For a pipe open at both ends, f = nv/2l n = 1, 2, ….. ⇒ 430 = \(\frac{n \times 340}{2 \times 0.2}\) ⇒ n = 0.51 Since n<1, resonance does not occur.

46.

A pipe 20 cm long Is closed at one end. Which harmonic mode of the pipe 1s resonantly excited by a 430 Hz source? Will the same source be In resonance with the pipe if both ends are open? (speed of sound In air Is 340 m s-1 ).

Answer»

Given :

l = 20 cm = 0.2 m

v = 340 ms^-1

f = 430 Hz

For a pipe closed at one end :

f = \(\frac{(2n-1)v}{4l}\) n = 1, 2, 3 ……

⇒ 430 = (2n-1)\(\frac{340}{4 \times 0.2}\)

⇒ n = 1

⇒ Resonance occurs only for first/fundamental mode of vibration. For a pipe open at both ends,

f = nv/2l n = 1, 2, …..

⇒ 430 = \(\frac{n \times 340}{2 \times 0.2}\)

⇒ n = 0.51

Since n<1, resonance does not occur.

Discussion

47.	A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s-1)
Answer» Length of the closed end pipe, l = 20 cm = 0.2 m Speed of sound, v = 340 ms^-1 The fundamental frequency, v = v/4l = {340}/{4 x 0.2} = 425 Hz This is the first harmonic. The second harmonic is 3v i.e., 1275 Hz, the third harmonic is 5v i.e., 2125 Hz etc. Therefore only the first harmonic of frequency 425 Hz is resonantly excited by a 430 Hz source. If both the ends of the pipe are open, the fundamental frequency is given by: v' = v/2l = {340}/{2 x 0.2} = 850 Hz The second, third, fourth ... etc. harmonics have frequencies 2v', 3v', 4v', ....i.e., 1700 Hz, 2550 Hz, 3400 Hz. etc. No harmonic can be excited by the 430 Hz source in this case.

47.

A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s-1)

Answer»

Length of the closed end pipe,

l = 20 cm = 0.2 m

Speed of sound, v = 340 ms^-1

The fundamental frequency, v = v/4l = {340}/{4 x 0.2}

= 425 Hz

This is the first harmonic. The second harmonic is 3v i.e., 1275 Hz, the third harmonic is 5v i.e., 2125 Hz etc. Therefore only the first harmonic of frequency 425 Hz is resonantly excited by a 430 Hz source.

If both the ends of the pipe are open, the fundamental frequency is given by:

v' = v/2l = {340}/{2 x 0.2} = 850 Hz

The second, third, fourth ... etc. harmonics have frequencies 2v', 3v', 4v', ....i.e., 1700 Hz, 2550 Hz, 3400 Hz. etc. No harmonic can be excited by the 430 Hz source in this case.

Discussion

48.	A wire stretched betweentwo right supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is `3.5 xx10^(-2) kg ` and its linear mass density is `4.0 xx10^(-2) kg m^(-1)`. What is (a) the speed of transverse wave on thestring . and (b) the tension in the string ?
Answer» Here, `v =45 Hz, mu = 3.5 xx 10^(-2) kg ` Mass `//` length `=mu =4.0 xx 10^(-2) kg //m` `l =(mu)/( mu )=(3.5 xx 10^(-2))/(4.0 xx 10^(-2)) = ( 7)/(8) ` As ` ( lambda)/( 2) = l = ( 7)/( 8) ` `:. lambda= ( 7)/(4)m =1.75 m ` (a) Thespeed of transverse wave ` v = v lambda= 45 xx 1.75 =78.75 m //s` (b) As `v =sqrt((T)/(mu))` `:. T= v^(2) xxmu = ( 78.75)^(2) xx4.0 xx 10^(-2)` `=248 .06 N `

Discussion

49.	Waves `y_(1) = Acos(0.5pix - 100pit)` and `y_(2)Acos(0.46pix - 92pit)` are travelling along x-axis. (Here `x` is in `m` and `t` is in second) (3) The number of times `y_(1) + y_(2) = 0` at `x = 0` in `1sec` isA. (a) `100`B. ( b) `46`C. ( c ) `192`D. (d) `96`
Answer» Correct Answer - D (d) The equations are `y_(1) = Acos (0.5 pix - 100 pit)` and `y_(2) = Acos (0.46 pix - 92 pit)` represent two prograssive wave travelling in the same direction with slight difference in the frequency. This will give the hpenomenon of beats. Comparing it with equation `y = A cos (kx - omegat)`, we get `omega_(1) = 100pu` rArr `2pif_(1) = 50 Hz` and `K_(1) = 0.5pi` rArr `(2pi)/(lambda_(1) = 0.5pi` rArr `lambda_(1) = 4m` Wave velocity `= lambda_(1)f_(1) = 200m//s` [Alternatively use `v = (omega)/(K)`] `omega^(2) = 2pif_(2) = 92pi` rArr `f_(2) = 46 Hz` Therefore beat frequency `= f_(1) - f_(2) = ` Hz` and `K_(2) = 0.46pi` rArr `(2pi)/(lambda_(2) = 0.46pi` rArr `lambda_(2) = (200)/(46)` Wave velocity `= (200)/(46) xx 46 = 200m//s`

Discussion

50.	Define the following terms. 1. Wave amplitude 2. Wave period 3. Wave frequency
Answer» 1. Wave amplitude (a): The maximum displacement of a particle of the medium from its mean position is called 1 wave amplitude. 2. Wave period (T): The time taken by a particle of the medium to complete one vibration or Wave period is the time during which one complete wave is set up in a medium. 3. Wave frequency (f): The number of vibrations completed by a particle of the medium in one second is called wave frequency or Wave frequency is the number of waves set up in the medium in one second.

50.

Define the following terms. 1. Wave amplitude 2. Wave period 3. Wave frequency

Answer»

1. Wave amplitude (a): The maximum displacement of a particle of the medium from its mean position is called 1 wave amplitude.

2. Wave period (T): The time taken by a particle of the medium to complete one vibration or Wave period is the time during which one complete wave is set up in a medium.

3. Wave frequency (f): The number of vibrations completed by a particle of the medium in one second is called wave frequency or Wave frequency is the number of waves set up in the medium in one second.

Discussion

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A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s-1)

A wire stretched betweentwo right supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is `3.5 xx10^(-2) kg ` and its linear mass density is `4.0 xx10^(-2) kg m^(-1)`. What is (a) the speed of transverse wave on thestring . and (b) the tension in the string ?

Waves `y_(1) = Acos(0.5pix - 100pit)` and `y_(2)Acos(0.46pix - 92pit)` are travelling along x-axis. (Here `x` is in `m` and `t` is in second) (3) The number of times `y_(1) + y_(2) = 0` at `x = 0` in `1sec` isA. (a) `100`B. ( b) `46`C. ( c ) `192`D. (d) `96`

Define the following terms. 1. Wave amplitude 2. Wave period 3. Wave frequency