320 + Interview Questions in SOUND-WAVES IN GENERAL AWARENESS Page 7 InterviewSolution

301.	A tuning fork of unknown frequency makes 5 beats per second with another tuning fork which can cause a closed organ pipe of length 40 cm to vibrate in its fundamental mode. The beat frequency decreases when the first tuning fork is slightly loaded with wax. Find its original frequency. The speed of sound in air is `320 ms^-1`
Answer» Correct Answer - B Given length of the closed organ pipe `I=40cm=40xx10^-2m` `V_(air)=320` So its frequency n=V/(4I)` =320/(4xx40xx10^-)=200Hz` As the tuning fork produces 5 bets with the closed pipe, its frequency must be 195 Hz or 205 Hz. Given that as it loaded its frequency decrease, so the frequency of tuning fork =205 Hz

Discussion

302.	A vibrating string produces 2 beats per secod when sounded with a turning fork of frequency `256Hz`. Slightly increasing the tension in the string produces 3 beats per second. The initial frequency of the string may have beenA. `253Hz`B. `254Hz`C. `258Hz`D. `259Hz`
Answer» Correct Answer - B::C::D

Discussion

303.	A closed organ pipe is sounded near a guitar, causing one of the string to vibrate with large ampulitude . We vary the tension of the string until we find the maximum amplitude. The string is `80%` aas long as the closed pipe. If both the pipe and the string vibrate at their fundamental frequency, calculate the ratio of the wave speed on the string to the speed of sound in air. s
Answer» Correct Answer - D `(nu)/(2(0.81)) = (nu_(2))/(4l) rArr (nu_(1))/(nu_(2)) = 0.4`

Discussion

304.	An organ pipe closed at one end has fundamental frequency of `500 Hz`. The maximum number of overtones generated by this pipe which a normal person can hear isA. `14`B. `13`C. `6`D. `9`
Answer» Correct Answer - C Audible frequency `= (20 - 20, 000 Hz)` `f_(1) = 1500 Hz, f_(max) = 20,000 Hz` `f_(max) = (2k+1)f_(1)`, `k`: constant of overtones `20000 = (2k+1) xx 1500` `k = 6.16` Maximum number of overtones `= 6`

Discussion

305.	A metal wire of diameter `1 mm` is held on two knife edges by a distance `50 cm`. The tension in the wire is `100 N`. The wire vibrating with its fundamental frequency and a vibrating tuning fork together produce `5 beats//s`. The tension in the wire is then reduced to `81 N`. When the two are excited, beats are heard at the same rate. Calculate (a) frequency of a fork and (b) the density of material of wire.
Answer» Correct Answer - (a) `95 Hz` , (b) `(40)/(pi)xx10^(3)kg//m^(3)`

Discussion

306.	Two identical piano wires have fundemental frequency of `600 vib//sec`, when kept under the same tension. What frectional increase in the tension of one wire will lead to the occurrence of six beats per second when both wires vibrate simuitaneously.
Answer» Correct Answer - `2%` `f = (1)/(lambda) sqrt((T)/(mu))` `lnf = -lnlambda + (1)/(2)lnT - (1)/(2)ln mu` `(1)/(f) (df)/(dT) = 0 + (1)/(2) (1)/(T) - 0` `(df)/(f) = (1)/(2)(dT)/(T)` `(dT)/(T) = (2df)/(f) = (2 xx 6)/(600) = (2)/(100)` `%` change `= (2)/(100) xx 100 = 2%`

Discussion

307.	The fundamental frequency of a vibrating organ pipe is 200 Hz.A. the first overtone is 400 Hz.B. The first overtone may be 400 HzC. The first overtone may be 600 HzD. 600 Hz is an overtone
Answer» Correct Answer - B::C::D

Discussion

308.	The third overtone of an open organ pipe of length `l_(0)` has the same frequency as the third overtone of a closed pipe of length `l_(c)`. The ratio `l_(0)//l_(c)` is equal toA. 2B. `3//2`C. `5//3`D. `8//7`
Answer» Correct Answer - D

Discussion

309.	The fundamental frequency of a vibrating organ pipe is `200 Hz`. (i) The first overtone is `400 Hz`. (ii) The first overtone may be `400 Hz`. (iii) The first overtone may be `600 Hz`. (iv) `600 Hz` is an overtone.A. (i), (iv)B. (i), (i), (iii)C. (iii), (iv)D. (ii), (iii), (iv)
Answer» Correct Answer - D Frequencies of closed pipe: `200, 600, 1000`, …. Frequencies of open pipe: `200, 400, 600, 800`, … (i) First overtone is `600 Hz` for closed pipe, `400 Hz` for open pipe, (i) is wrong (iv) `600 Hz` is first overtone of closed pipe, second overtone of open pipe, (iv) is `O.K`.

Discussion

310.	An open organ pipe has a fundamental frequency of `300 H_(Z)` . The first overtone of a closed organ pipe has the same frequency as the first overtone of this open pipe . How long is each pipe ? (Speed of sound in air = `330 m//s`)
Answer» Correct Answer - C Fundamental freqency of an open organ pipe, `f_(1)= (nu)/(2l_(0)) rArr l_(0) = (nu)/(2f_(1)) = (330)/(2xx300)` = `0.55m` = `55 cm` Given, first overtone of closed organ pipe = first overtone of open organ pipe Hence, `3((nu)/(4l_(C))) = 2((nu)/(2l_(0)))` `:. l_(C) = (3)/(4) l_(0) = ((3)/(4)) (0.55)` = `0.4125 m` = `41. 25 cm`

Discussion

311.	The first overtone frequency of a closed organ pipe `P_1` is equal to the fundamental frequency of an open organ pipe `P_2`. If the length of the pipe `P_1` is 30 cm, what will be the length of` P_2`?
Answer» Correct Answer - B::C According to the questions `f_1` first overtone of a closed organn pipe `P_1=(3v)/(4I)=((3xxV)/(4xx30)) `=f_2=` fundamental frequency of a open organ pipe `P_2=(V/(2I_2))` `Here given ((3xxV)/(4xx30))=V/(2I_2)` `rarr I_2=20cm` `:. length of the pipe `P_2` will be 20 cm.

Discussion

312.	the fundamental frequency of a closed organ pipe is `50 Hz`. The frequency of the second overtone isA. `100 Hz`B. `150 Hz`C. `200 Hz`D. `250 Hz`
Answer» Correct Answer - D Second overtone `= (2k+1)f_(1)` `= (2 xx 2 + 1) xx 50 =250 Hz`

Discussion

313.	If `lambda_(1), lambda_(2), lambda_(3)` are the wavelengths of the waves giving resonance in the fundamental, first and second overtone modes respecively in a open organ pipe, then the ratio of the wavelengths `lambda_(1) : lambda_(2) : lambda_(3)`, is :A. `1 : 2 : 3`B. `1 : 3 : 5`C. `1 : 1//2 : 1//3`D. `1 : 1//3 : 1//5`
Answer» Correct Answer - C `n_(1) : n_(2) : n_(3) = 1 : 2 : 3` `(V)/(lambda_(1)) : (V)/(lambda_(2)) : (V)/(lambda_(3)) = 1 : 2 : 3` `lambda_(1) : lambda_(2) : lambda_(3) = 1 : (1)/(2) : (1)/(3)`

Discussion

314.	Two sources of sound placed close to each other are wmitting progressive waves given by `y_1=4sin600pit` and `y_2=5sin608pit`. An observer located near these two sources of sound will hear:A. `4` beats per second with intensity ratio `25 : 16` between waxing and waningB. `8` beats per second with intensity ratio `25 : 16` between waxing and waningC. `8` beats per second with intensity ratio `81 : 1` between waxing and waningD. `4` beats per second with intensity ratio `81 : 1` between waxing and waning
Answer» Correct Answer - D `y_(1) = 4 sin 600 pi t = A_(1) sin 2 pi f_(1) t` `A_(1) = 4, f_(1) = 300 Hz` `y_(2) = 5 sin 608 pi t = A_(2) sin 2pi f_(2) t` `A_(2) = 5, f_(2) = 304 Hz` Beat frequency `= f_(1)~f_(2) = 4 Hz` `(I_(max))/(I_(min)) = ((A_(1)+A_(2))/(A_(1)~A_(2)))^(2) = ((4+5)/(4~5))^(2) = (81)/(1)`

Discussion

315.	A police car moving at `22m//s`, chases motorcyclist. The police man sounds his horn at `176 Hz`, while both of them move towards a ststionary siren of frequency `165 Hz`. Calculate the speed of the motorcycle, if it is given that he does not observes any beats. A. 33 m/sB. 22 m/sC. zeroD. 11 m/s
Answer» Correct Answer - B The motorcyclist observes no beats. So the apparent frequency observed by him from the sources must be equal. `f_(1)`=Frequency recorded by motorcyclist from police car. `f_(2)`=Frequency recored by motorcyclist from stationary siren. `:. 176((330-upsilon)/(330-22))=165((330+upsilon)/330)` Solving this equation we get, `upsilon=22 m//s`

Discussion

316.	A police car moving at 22 m/s, chases a motorcylist. The police man sounds his horn at 176 Hz, while both of them move towards a stationary siren of frequency 165 Hz. Calculate the speed of the motorcycle, if it is given that he does not observes any beats A. `33 m//s`B. `22 m//s`C. zeroD. `11 m//s`
Answer» Correct Answer - B

Discussion

317.	A string of length `20 cm` and linear mass density `0.40 g//cm` is fixed at both ends and is kept under a tension of `16 N.A` wave pulse is produced at `t=0` nearj an end as shown in figure which travels towards the other end. when will the string have the shape shown in the figure again? `(inxx10^(-2)s)` A. `0.05s`B. `0.1s`C. `0.2s`D. `0.4s`s
Answer» Correct Answer - B

Discussion

318.	Four sources of sound each of sound level `10 dB` are sounded together in phase , the resultant intensity level will be `(log _(10) 2 = 0.3)`A. `40 dB`B. `26 dB`C. `22 dB`D. `13 dB`
Answer» Correct Answer - C Resulant ampulitude will become `4` times. Therefore , resultant intensity is `16` times `L_(2) - L_(1) = log_(10) (I_(2))/(I_(1))` or `I_(2) - 10 = 10 log _(10) (16)` or `L_(2) = 22 dB`

Discussion

319.	Two sources of sound of the same frequency produce sound intensities `I` and `4I` at a point `P` when used individually. If they are used together such that the sounds from them reach `P` with a phase differenceof `2pi//3`, the intensity at `P` will beA. `2I`B. `3I`C. `4I`D. `5I`
Answer» Correct Answer - B

Discussion

320.	A window whose area is `2 m^(2)` opens on a street where the street noise results at the window an intensity level of `60 dB`. How much acoustic power energy from the street will it collect in a day?
Answer» Correct Answer - A::B::C By definition sound level `= 10 log.(I)/(I_(o)) = 60` or `(I)/(I_(o)) = 10^(6)` rArr `I = 10^(-12) xx 10^(6) = 1mu W//m^(2)` power entering the room `= 1 xx10^(-6) xx 2 = 2mu W` Energy collected in a day `= 2 xx10^(-6) xx 86400` `= 0.173 J`

Discussion

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A vibrating string produces 2 beats per secod when sounded with a turning fork of frequency `256Hz`. Slightly increasing the tension in the string produces 3 beats per second. The initial frequency of the string may have beenA. `253Hz`B. `254Hz`C. `258Hz`D. `259Hz`

An organ pipe closed at one end has fundamental frequency of `500 Hz`. The maximum number of overtones generated by this pipe which a normal person can hear isA. `14`B. `13`C. `6`D. `9`

Two identical piano wires have fundemental frequency of `600 vib//sec`, when kept under the same tension. What frectional increase in the tension of one wire will lead to the occurrence of six beats per second when both wires vibrate simuitaneously.

The fundamental frequency of a vibrating organ pipe is 200 Hz.A. the first overtone is 400 Hz.B. The first overtone may be 400 HzC. The first overtone may be 600 HzD. 600 Hz is an overtone

The third overtone of an open organ pipe of length `l_(0)` has the same frequency as the third overtone of a closed pipe of length `l_(c)`. The ratio `l_(0)//l_(c)` is equal toA. 2B. `3//2`C. `5//3`D. `8//7`

The fundamental frequency of a vibrating organ pipe is `200 Hz`. (i) The first overtone is `400 Hz`. (ii) The first overtone may be `400 Hz`. (iii) The first overtone may be `600 Hz`. (iv) `600 Hz` is an overtone.A. (i), (iv)B. (i), (i), (iii)C. (iii), (iv)D. (ii), (iii), (iv)

An open organ pipe has a fundamental frequency of `300 H_(Z)` . The first overtone of a closed organ pipe has the same frequency as the first overtone of this open pipe . How long is each pipe ? (Speed of sound in air = `330 m//s`)

The first overtone frequency of a closed organ pipe `P_1` is equal to the fundamental frequency of an open organ pipe `P_2`. If the length of the pipe `P_1` is 30 cm, what will be the length of` P_2`?

the fundamental frequency of a closed organ pipe is `50 Hz`. The frequency of the second overtone isA. `100 Hz`B. `150 Hz`C. `200 Hz`D. `250 Hz`

A police car moving at `22m//s`, chases motorcyclist. The police man sounds his horn at `176 Hz`, while both of them move towards a ststionary siren of frequency `165 Hz`. Calculate the speed of the motorcycle, if it is given that he does not observes any beats. A. 33 m/sB. 22 m/sC. zeroD. 11 m/s

Four sources of sound each of sound level `10 dB` are sounded together in phase , the resultant intensity level will be `(log _(10) 2 = 0.3)`A. `40 dB`B. `26 dB`C. `22 dB`D. `13 dB`

Two sources of sound of the same frequency produce sound intensities `I` and `4I` at a point `P` when used individually. If they are used together such that the sounds from them reach `P` with a phase differenceof `2pi//3`, the intensity at `P` will beA. `2I`B. `3I`C. `4I`D. `5I`

A window whose area is `2 m^(2)` opens on a street where the street noise results at the window an intensity level of `60 dB`. How much acoustic power energy from the street will it collect in a day?