37 + Interview Questions in WAVE-MOTION-AND-WAVES-ON-A-STRING IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Two waves of equal amplitude A, and equal frequency travel in the same direction in a medium. The amplitude of the resultant wave is
Answer» Correct Answer - D

Discussion

2.	Two periodic waves of amplitudes `A_1 and A_2` pass though a region. If `A_1gtA_2`, the difference in the maximum and minimum resultant amplitude possible isA. `2A_1`B. `2A_2`C. `A_1+A_2`D. `A_1-A_2`
Answer» Correct Answer - B

Discussion

3.	A transverse wave travels along the Z-axis. The particles of the medium must moveA. along the Z-axisB. along the X-axisC. along the Y-axisD. in the XY-plane
Answer» Correct Answer - D

Discussion

4.	A 660 Hz tuning fork sets up vibration in a string clamped at both ends. The wave speed for a transverse wave on this string is `1220 m s^-1` and the string vibrates in three loops. (a) Find the length of the string. (b) If the maximum amplitude of a particle is 0.5 cm, write a suitable equation describing the motion.
Answer» Correct Answer - a) 50cm b) `cos(0.06picm^-1)sin((1320pis^-1)t)` Frequency of the tunning fork, `f=660Hz` Wave speed `v=220m/s` `rarr `lamda`=V/f=220/660=1/3m` No.of loops = 3 a. So, L.`f=(3/2)v` `rarr L.660=(3/2)xx220` ` L=1/2m=50cm` b. the equation of resultant statioN/Ary wave is given by, `y=2Acos((2pix)/lamda)sin((2pivt)/lamda)` `rarr y=(0.5)cos((2pix)/(1/3m))sin((2pixx220xxt)/lamda)` `rarr y=(0.5cm)` `cos(6pixm^-1)sin(1320pis^-1t)` `rarr y=(0.5cm)` `cos(0.06picm^-1)sin(1320pis^-1t)`

Discussion

5.	Figure shows a plot of the transverse displacements of the particles of a string at t = 0 through which a travelling wave is passing in the positive x-direction. The wave speed is `20 cms^-1`. Find (a) the amplitude, (b) the wavelength, (c) the wave number and (d) the frequency of the wave..
Answer» Correct Answer - A::B::C::D Given that `v=20cm/s` a. amplitude `A=1mm` b. `Wavelength lamda=4cm` c. Wave number `n=(2pi)/lamda` `=((2xx3.14))/4` `=1.57cm^-1` `(wave number =k)` d. Frequencey `f=1/T=20/4=5Hz` `(where time period T=lamada/v)`

Discussion

6.	Two waves are smultaneously passing through a string. The equation of the waves are given by `y_1=A_1sink(x-vt)` and `y_2=A_2sink(x=vt+x_0)` where the wave number `k=6.28 cm^-1 and x_0=o1.50c. The ampitudes of `A_1=5.0mm and A_2=4.0mm. find the phase difference between the waves and the amplitude of the resulting wave.
Answer» The phase of the first wave is `k(x-vt)` and of the second is `k(x-vt+x_0)` The phase difference is therefore, `delta=k x_0=(6.28cm^-1)(1.50cm)=2pixx1.5=3pi` The waves satisfy the condition of destructive interference. The amplitude of the resulting wave is given by `\|A_1-A_2\|=5.0mm-4.0mm=1.0mm`.

Discussion

7.	A pulse travelling on a string is represented by the function y=a^3/((x-vt)^2+a^2)` where a=5 mm and v=20 cms^-1. Sketch the shape of the string at t=0, 1 s and 2s. Take x=0 in the middle of the string.
Answer» Correct Answer - A The pulse is given by `y=[((a)^3)/([(x-vt)^2+a^2)])]` `a=5mm` `=0.5cm` `v=20cm/s` At `t=0s, y=a^3/((x^2+a^2))` The graph between y and x can be plotted by talking different vaues of x. (left as excercse for the studen). ltbr.gt similarly at t=1 s, `y=a^3/({(x-v)^2+a^2])` and at t=2s, y=a^3/({(x-2v)^2+a^2})`

Discussion

8.	A string of length 40 cm and weighing 10 g is attached to a spring at one end and to a fixed wall at the other end. The spring has a spring constant of `160 N m^-1` and is stretched by 1-0 cm. If a wave pulse is produced on the string near the wall, how much time will it take to reach the spring ?
Answer» L=40 cm, mass=10gm ………mass per unit length `=10/40=1/4(gm/cm)` `Spring costant=k=160N/m` Deflection =x=1cm` `=0.01m` `rarr T=kx=160x0.01` `=1.6N=6xx10^4 dyne` Again `v=sqrt((T/m)0=sqrt(((16x10^4)/(1/4)))` `=8xx10^2cm/s=800cm/s` `:. time taken by the pulse to reach the spring. `t=40/800=1/20=0.05sec`

Discussion

9.	Figure shows a string stretched by a block going over a pulley. The string vibrates in its tenth harmonic in unison with a particular tuning fork. When a beaker containing water is brought under the block so that the block is completely dipped into the beaker, the string vibrates in its eleventh harmonic. Find the density of the material of the block. .
Answer» Correct Answer - 5.8x10^3kg/m^2` Let `rhorarr` density of the block `weight= rhoVg` Where V=volume of block. The same tunning fork resonates with string in the two cases. `f_10=10/(2L) sqrt(T/m)` `=10/(2L) sqrt(rhoVg)/m` `(mrarr `mass per unit length of string) When the block is immersed in water `f_11=11/(2L) sqrt((T-rho_wVg)/m)` `=11/(2L) sqrt((rho-rho_w)Vg)/m` As the for of tunning fork is same `f_10=f_11=10/(2L) sqrt((rhoVg)/m)` `=11/(2L)sqrt(((rho-rho_w)Vg)/m)` `rarr 10/(2L)sqrt((rho-rho_w)/m)=(rho-1)/rho= 100/121` `(because rhow=1(gm)/cc)` `rarr 100 rho=121rho-121` `rarr rho=121/21` `=5.8(gm)/cc` `=5.8(gm)/cc` `=5.8x10^3(kg)/m^3`

Discussion

10.	A wave pulse is travelling on a string with a speed v towards the positive X-axis. The shape of the string at t = 0 is given by `g(x) = A sin(x /a)`, where A and a are constants. (a) What are the dimensions of A and a ? (b) Write the equation of the wave for a general time 1, if the wave speed is v.
Answer» Correct Answer - A::B At `t=0, g(x)=Asin(x/a)` a. [M^0L^1T^0]=[L]` `a=[M^0L^1T^0]=[L]` b. wave speed =v `:. Time period T=a/v` `(a=wave length =lamda)` `:.` General of wave `y=Asin{(x/a-t/((a/v))}` `=AsinP((x-vt))/a}`

Discussion

11.	A heavy string is tied at one end to a movable support and to a light thread at the other end as shown in figure. The thread goes over a fixed pulley and supports a weight to produce a tension. The lowest frequency with which the heavy string resonates is 120 Hz. If the movable support is pushed to the right by 10 cm so that the joint is placed on the pulley, what will be the minimum frequency at which the heavy string can resonate ? .
Answer» Correct Answer - 240 Hz Initiallly because the end A is free and antinode will be formed. `So, I=lambda_1/4` Again if the movable support is pushed to right by 10 m, so that the joint is placed on the poulley, node will be formed there So, `I=(lambda)/2` Since the tension remains same in both the case, velocity remains same as the wavelength is reduced by half the frequency will become twice as that of 120 Hz i.e. 240 Hz.

Discussion

12.	A string of linear mass density `0.5 g cm^-1` and a total length 30 cm is tied to a fixed wall at one end and to a frictionless ring at the other end. The ring can move on a vertical rod. A wave pulse is produced on the string which moves towards the ring at a speed of `20 cm s^-1`. The pulse is symmetric about its maximum which is located at a distance of 20 cm from the end joined to the ring. (a) Assuming that the wave is reflected from the ends without loss of energy, find the time taken by the string to regain its shape. (b) The shape of the string changes periodically with time. Find this time period. (c) What is the tension in the string ?
Answer» Correct Answer - A::B::C The crest reflects as a crest here as the wave is travelling from denser to rarer medium rarr Phase chnge =0 a. To regain shape, after travelled by the wave `S=20+20=40cm` Wave speed v=20 m/s `rarr Time =s/v=40/20=2sec` b. The wave regains its shape after travelling a period distance `=2xx30=60cm` `:. time period =60/20=3sec`. c. Frequency n=(1/2sec^2)` `=1/(2l)sqrt((T/m))` m=mass per unit length `=0.5gm/cm` `rarr 1/3=1/((2xx30))sqrt((T/0.5)0` `rarr T=400xx0.5` = 200 dyne `=2xx 10^(-3) Newton`

Discussion

13.	A sonometer wire having a length of 1.50 m between the bridges vibrates in its second harmonic in resonance with a tuning fork of frequency 256 Hz. What is the speed of the transverse wave on the wire ?
Answer» Correct Answer - A::C::D First harmonit be `f_0`. Second harmonic be `f_1` `rarr f_1=2f_0` `f_1=256Hz` `:.1`st harmonic or fundamental frequency `=f_0=f_1/2=256/2=128Hz` `lamda/2=1.5m` `rarr lamda=3m` [when fundamental wave is produced] `rarr Wave speed =V` `=f_0lamda=128xx3=384m/s`

Discussion

14.	A 40 cm wire having a mass of 3.2 g is stretched between two fixed supports 40.05 cm apart. In its fundamental mode, the wire vibrates at 220 Hz. If the area of cross section of the wire is `1.0 mm^2`, find its Young modulus.
Answer» Correct Answer - 1.985xx10^11N/m^2 `L=40cm=.4m` mass=3.2g=3.2x10^-3kg` `:. Mass per unit length `m=(3.2xx10^-3)/0.4=8xx10^-3(kg)/m` Change in length `/_\L=40.05=80xx10^-3(kg)/m` Strain` =(/_\L)/L=((0.05xx10^-2))/0.4` `=0.125xx10^-2` `f=220Hz` `f=1/(2L)sqrt(T/m)` `=1/(2xx(0.4005))` `sqrt((T/(8xx10^-3)))` `rarr 220xx220=[1/((0.801)^2)]xxTxx{10^3/8)` `rarr Txx1000=220xx220xx0.641xx0.8` `rarr T=248.19N` Strain =`248.19/1(1mm^2)` Strain =`248.19/(1mm^2)` `=248.19/10^-6=248.19xx10^6` `Y`=stress/strain `=((248.19xx10^6))/((0.125xx10^-2)) =1985.52xx10^8 =1.985xx10^11N/m^2`

Discussion

15.	A string of length 20 cm and linear mass density `0.40 g cm^-1` is fixed at both ends and is kept under a tension of 16 N. A wave pulse is produced at t = 0 near an end as shown in figure, which travels towards the other end. (a) When will the string have the shape shown in the figure again ? (b) Sketch the shape of the string at a time half of that found in part (a).
Answer» Correct Answer - A::B a. Velcity of the wave `v=sqrt((T/m))` `sqrt((16x1^5)/0.4)` `=2000cm/sec` `:. Time taken to reach to the other end `=20/2000=0.01sec` Time taken to see the pulse again in the origiN/Al position `=0.01xx2=0.02sec` b. At t=0.01s, then will be a trough at the right end as it is reflected.

Discussion

16.	A 2.00 m-long rope, having a mass of 80 g, is fixed at one end and is tied to a light string at the other end. The tension in the string is 256 N. (a) Find the frequencies of the fundamental and the first two overtones. (b) Find the wavelength in the fundamental and the first two overtones.
Answer» Correct Answer - A::B::C L=length of rope =2m, M=mass=80gm =0.08kg mass per unit length =m `=0.08/2=0.04(kg)/m` Tension , `T=256N` Velocity , `V=sqrt((T/m))` `=sqrt((25600/4))=160/2=80m/s` For fundamental frequency `I=lamda/4` `rarr lamda=4I=4xx2=8m` `rarr f=80/8=10Hz` a. therefore, the frequency of 1st two overtones are 1st overtone =3f=30 Hz 2nd overtone =5f=50Hz. b. `lamda=4I=4xx2=8m` `:. lamda_1=V/(f_1) =80/30=2.67m` `lamda_2=V/(f_2) =80/50=1.6m` So the wavelength are 8, 2.67m and 1.6m respectively.

Discussion

17.	Figure shows an aluminium wire of length 60 cm joined to a steel wire of length 80 cm and stretched between two fixed supports. The tension produced is 40 N. The cross-sectional area of the steel wire is .`1.0 mm^2 ` and that of the aluminium wire is 3.0 mm 2. What could be the minimum frequency of a tuning fork which can produce standing waves in the system with the joint as a node ? The density of aluminium is `2.6 g cm^-3`and that of steel is ` 7.8gcm^-3`
Answer» Correct Answer - A `P_s=7.8(gm)/cm^3` `P_A=2.6(gm)/cm^3` `m_s=P_sA_s` `=7.8xx10^-2(gm)/cm` (m=mass per unit length) `7.8xx10^-3kg/m` `m_A=P_A` `=21.6xx10^-2xx3(gm)/cm` `=7.8xx10^-2(gm)/cm` `=7.8xx10^-3(kg)/m` A node is always placed in the joint. Since aluminium and steel rod has same mass per unit lenght, velocity of wave in both of them is same `rarr v=sqrt((T/m))` `=sqrt({40/((7.8xx10^-3))})` `=sqrt((((4xx10^4)))/7.8)` `=71.6m/s` For minimum frequency there would be maximum wavelength. for maximu wavelength minimum no. of loops are to be produced. `:.` Maximum distance of a loop =20cm `rarr` wavelength =`lamda=2xx20` `=40cm=0.4m`=180Hz`

Discussion

18.	A steel wire of mass 4.0 g and length 80 cm is fixed at the two ends. The tension in the wire is 50 N. Find the frequency and wavelength of the fourth harmonic of the fundamental.
Answer» Correct Answer - frequency = 250 and `lambda = 40cm` m=`(4/80)g/(cm)` `=0.005(kg)/m` `T=50N` `L=80cm=0.8m` `v=sqrt((T/m))` `=sqrt((50/0.005))=100m/s` Fundamental frequency `=f=1/(2L)sqrt((T/m))` `={1/(2xx0.8)xxsqrt((50/0.005))}` `=100/((2xx0.8))` `=100/1.6=62.Hz` First `=62.5Hz` `f_4`=Frequency of forth harmonic `=4f_0=f_3=62.5xx4` `=250Hz` `v=f_4 lambda_4` `rarr (v/f)=100/250` `=0.4m=40cm`

Discussion

19.	A guitar string is 90 cm long and has a fundamental frequency of 124 Hz. Where should it be pressed to produce a fundamatal frequecy of 186 Hz?
Answer» The fundamental frequency of a string fixed at both ends is given by `v =(1)/(2L) sqrt((F)/(mu))` As F and mu are fixed, `(v_1)/(v_2) = (L_2)/(L_1)` or , `L_2 = (v_1)/(v_2) L_1 = (124 Hz)/(186 Hz) (90 cm) = 60 cm.` Thus, the string should be pressed at 60 cm from an end.

Discussion

20.	A wire, fixed at both ends is seen to vibrate at a resonant frequency of 240 Hz and also at 320 Hz. (a) What could be the maximum value of the fundamental frequency ? (b) If transverse waves can travel on this string at a speed of 40 m s t, what is its length ?
Answer» Correct Answer - 0.25m This wire makes a resonant frequency of 240 Hz and 320 Hz. The fundamental frequency of the wire must be factor of 240 Hz and 320 Hz. a So, the maximum value of fudamental frequency is 80 Hz. b. Wave speed `v=40 m/s` `rarr Lxx80=(1/2)xx40` `rarr L=1/4m=0.25m`

Discussion

21.	A string, fixed at both ends, vibrates in a resonant mode with a separation of 2.0 cm between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to 1.6 cm. Find the length of the string.
Answer» Correct Answer - C Let there be n loops in the 1st case `rarr` Length of the wire` `L =((n`lamda_1`)/2) (`lamda_1`=`2xx2=4cm`) `rarr` Length of the wire `L_l={(n+1)lamda_2/2}` `(lamda_2=2xx(1.6)=3.2cm)` `rarr (nlamda_1)/2=(n+1)lamda_2/2` `rarr nxx4=(n+1)(3.2)` `rarr 4n-(3.2)n=3.2` `rarr 0.8n=3.2` `rarr n=4` `:. ` Length of the string `L=(n lamda_1)/2=((4xx4))/2=8 cm`

Discussion

22.	The average power transmitted through a given point on a string supporting a sine wave is 0.20 W when the amplitude of the wave is 2.0 mm. What power will be transmitted through this point if the amplitude is increased to 3.0 mm.
Answer» Other things remaining the same the power transmitted is proportioN/Al to the square of the amplitude Thus, `P_2/P_1=A_2^2/A_1^2` or, `P_2/(0.20W)=9/4=2.25` or, `P_2=2.25xx0.2W=0.45W`

Discussion

23.	A particle on a stretched string supporting a travelling wave, takes 5.0 ms to move from its mean position to the extreme position. The distance between two consecutive particles, which are at their mean positions, is 2.0 cm. Find the frequency, the wavelength and the wave speed.
Answer» Correct Answer - A::B::C::D Time period `T=4xx5ms` `=20x106-3=2xx10^-2s` `lamda=2xx2cm=4cm` `Frequency f=1/T` `=1/((2x10^-2))=50s^-1` `=50Hz` `Wave speed, lamdaf=4xx10^-2xx50m/s` `=200xx10^-2m/s=2m/s`

Discussion

24.	The equation of a wave travelling on a string is `y=(0.10mm)sin[3.14m^-1)x+(314s^-1)t]`. (a) In which direction does the wave travel ? (b) Find the wave speed, the wavelength and the frequency of the wave. (c) What is the maximum displacement and the maximum speed of a portion of the string ?
Answer» Correct Answer - A::B::C::D The equation of the wave is given by `y=[(01 mm)sin(314m^-1)x+(31.4s^-1)t]` `y=rsin{((2pix)/lamda)+omegat}` a. Negative x -direction ltbr. b.`k=31.4m^-1` `rarr (2pi)/lamda=31.4` `rarr lamda=(2pi)/31.4=0.2m=20cm` Again `w=314s^-1` `rarr 2pif=314` `rarr f=314/(2pi)` `=314/(2xx3.14)` `=50sec^1=50Hz` wave speed `v=lamdaf=20xx50` `=1000cm/s` `c. Max displacement =0.10mm `Max. velocity =`aw=0.1xx10^-1xx314` `=3.14cm/sec`

Discussion

25.	Mark out the correct optionsA. the energy of any smal part of a string remains constnt in a traveling waveB. the energy of any small part of a string remains constant in a standing waveC. the energies of all the smal parts of equal length are equal in a traveling waveD. The energies of all the small parts of equal length are equal in a standing wave
Answer» Correct Answer - B

Discussion

26.	In a stationary waveA. all the particles of the medium vibrate in phaseB. all the antinodes vibrates in phaseC. the alternate antinodes vibrate in phaseD. all the particles between consecutive nodes vibrate in phase.
Answer» Correct Answer - C::D

Discussion

27.	A standing wave is produced on a string on a string clamped at one end and free at the other. The length of the stringA. must be an integral multiple `lamda/4`B. must be an integral multiple of `lamda/2`C. must be an integral multiple of `lamda`D. may be an integral multiple of `lamda/2`
Answer» Correct Answer - A

Discussion

28.	A wave is represented by the equation `y=(0.001mm)sin[(50s^-10t+(2.0m^-1)x]`A. The wave velocity `=100 ms^-1`B. The wavelength =2.0mC. the frequencey `=25/piHz`D. The amplitude `=0.001mm
Answer» Correct Answer - C::D

Discussion

29.	A wave moving in a gasA. must be longitudinalB. may be longitudinalC. must be transverseD. may be transverse.
Answer» Correct Answer - A

Discussion

30.	Two particles A and B have a phase diference of `pi` when a sine wave passes through the reginA. A oscillates at half the frequency of BB. A and B move in opposite directionsC. A and B must be separated by half of the wavelength.D. The displacements at A and B have equal magnitudes
Answer» Correct Answer - B::D

Discussion

31.	Two blocks each having a mass of 3.2 kg are connected by a wire CD and the system is suspended from the ceiling by another wire AB . The linear mass density of the wire AB is 10 g m4 and that of CD is `8 g m^-1.` Find the speed of a transverse wave pulse produced in AB and in CD.
Answer» Correct Answer - A::C::D `m_1=m_2=3.2kg` `Mass per unit length of AB `=10gm/m` `=0.01kg/m` `Mass per unit length of CD `=8gm/m` `=0.008kg/m` `For the string CD, `rarr v=sqrt((T/m))` `=sqrt((3.2x10)/0.008)` `=sqrt((32x10^3)/8)` `=2xx10sqrt10` `=20xx3.14=63ms` For the string AB, `T=2x3.2g=6.4xxg` `=64N` `rarr v=sqrt((T/m))` `=sqrt((64/0.01))=sqrt(6400)`

Discussion

32.	The length of the wire shown in figure between the pulley is 1.5 m and its mass is 12.0 g. Find the frequency of vibration with which the wire vibrates in two loops leaving the middle point of the wire between the pulleys at rest.
Answer» Correct Answer - 70Hz `L=1.5m, mass=12gm` Mass per unit length `m=12/15g/m` `=8xx10^-3(kg)/m` `T=9xxg` `=90N,lamda=1.5m` `f_1=2/(2L)sqrt((T/m))` [for second harmonic two loops are produced] `f_1=2f_0=1/1.5 sqrt((90/8xx10^-3))` =`((1.6.06))/(1.5)` `=70.7Hz~~70Hz`

Discussion

33.	Which of the following equations represents as wave travelling along Y-axis?A. `x=Asin(ky-omegat)`B. `y=Asin(kx-omegat)`C. `y=Asinkycosomegat`D. `y=Acoskysinomegat`.
Answer» Correct Answer - A

Discussion

34.	A sine wave is travelling in a medium. A particular partile has zero displacement at a certain instant. The particle closest to it having zero displacement is at a distanceA. `lamda/4`B. `lamda/3`C. `lamda/2`D. `lamda`
Answer» Correct Answer - C

Discussion

35.	A wave travels along the positive x-direction with a speed of `20 ms^-1`. The amplitude of the wave is 0.20 cm and the wavelength 2.0 cm. (a) Write a suitable wave equation which describes this wave. (b) What is the displacement and velocity of the particle at x = 2.0 cm at time t = 0 according to the wave equation written ? Can you get different values of this quantity if the wave equation is written in a different fashion ?
Answer» Correct Answer - A::B::C::D Wave speed v=20 m/s `A=0.20 cm` `:. K=(2pi)/lamda=(2pi)/2=picm^-1` `T=lamda/v=2/2000` `=1/1000 se =10^-3sec` `omega=(2pi)/T` `=2pixx10^3sec^-1` So, the wave equation is `:. Y=(0.2cm)sin` `[(picm^-1)x(2pixx10^-3sec^-1)]` b. At `x=2cm and t=0, y=(0.2)cmsin(pi)=0` `:. v=romegacospix` `=0.2x2000pixxcos2pi` `=400pi` `=400picm/s=4pim/s`

Discussion

36.	Velocity of sound in air is 332 `ms^-1`. Its velocity in vacuum will beA. `gt332ms^-1`B. `=332ms^-1`C. `lt332ms^-1`D. meaningless
Answer» Correct Answer - D

Discussion

37.	Two waves passing through a region are represented by `y=(1.0cm) sin [(3.14 cm^(-1))x - (157s^(-1) x - (157s^(-1))t]` and `y = (1.5 cm) sin [(1.57 cm^(-1))x- (314 s^(-1))t].` Find the displacement of the particle at x = 4.5 cm at time t = 5.0 ms.
Answer» According to the principle of superpositon each wave produces its distrubance independent of the other and the resultant distrubance is eqal to the vector sum of the indicidual disturbances, The displacemts of the particle at x = 4.5 cm at time t= 5.0 ms due to the two waves are. `y_1 = (1.0cm) sin [(3.14 cm^(-1)) (4.5 cm)` -(157s ^(-1) (5.0xx10^(-9)s)]` `=(1.0cm)sin[4.5 pi - (pi)/(4)]` `=(1.0 cm) sin [4pi + pi//4] = (1.0cm)/(sqrt2)` and `y_2 = (1.5 cm) sin [1.57cm^(-1)) (4.5 cm)` `- (314s^(-1)) (5.0xx10^(-3)s)]` `=(1.5 cm) sin[ 2.25 pi - (pi)/(2)]` `=(1.5 cm) sin[2pi - pi/4]` `= - (1.5cm) sin(pi)/(4) = -(1.5cm)/(sqrt2)` The net displacment is `y =y_1 +y_2 = (-0.5cm)/(sqrt2) = -0.35cm.`

Discussion

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A pulse travelling on a string is represented by the function y=a^3/((x-vt)^2+a^2)` where a=5 mm and v=20 cms^-1. Sketch the shape of the string at t=0, 1 s and 2s. Take x=0 in the middle of the string.

A string of length 40 cm and weighing 10 g is attached to a spring at one end and to a fixed wall at the other end. The spring has a spring constant of `160 N m^-1` and is stretched by 1-0 cm. If a wave pulse is produced on the string near the wall, how much time will it take to reach the spring ?

A sonometer wire having a length of 1.50 m between the bridges vibrates in its second harmonic in resonance with a tuning fork of frequency 256 Hz. What is the speed of the transverse wave on the wire ?

A 40 cm wire having a mass of 3.2 g is stretched between two fixed supports 40.05 cm apart. In its fundamental mode, the wire vibrates at 220 Hz. If the area of cross section of the wire is `1.0 mm^2`, find its Young modulus.

A 2.00 m-long rope, having a mass of 80 g, is fixed at one end and is tied to a light string at the other end. The tension in the string is 256 N. (a) Find the frequencies of the fundamental and the first two overtones. (b) Find the wavelength in the fundamental and the first two overtones.

A steel wire of mass 4.0 g and length 80 cm is fixed at the two ends. The tension in the wire is 50 N. Find the frequency and wavelength of the fourth harmonic of the fundamental.

A guitar string is 90 cm long and has a fundamental frequency of 124 Hz. Where should it be pressed to produce a fundamatal frequecy of 186 Hz?

A wire, fixed at both ends is seen to vibrate at a resonant frequency of 240 Hz and also at 320 Hz. (a) What could be the maximum value of the fundamental frequency ? (b) If transverse waves can travel on this string at a speed of 40 m s t, what is its length ?

A string, fixed at both ends, vibrates in a resonant mode with a separation of 2.0 cm between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to 1.6 cm. Find the length of the string.

The average power transmitted through a given point on a string supporting a sine wave is 0.20 W when the amplitude of the wave is 2.0 mm. What power will be transmitted through this point if the amplitude is increased to 3.0 mm.

A particle on a stretched string supporting a travelling wave, takes 5.0 ms to move from its mean position to the extreme position. The distance between two consecutive particles, which are at their mean positions, is 2.0 cm. Find the frequency, the wavelength and the wave speed.

The equation of a wave travelling on a string is `y=(0.10mm)sin[3.14m^-1)x+(314s^-1)t]`. (a) In which direction does the wave travel ? (b) Find the wave speed, the wavelength and the frequency of the wave. (c) What is the maximum displacement and the maximum speed of a portion of the string ?

In a stationary waveA. all the particles of the medium vibrate in phaseB. all the antinodes vibrates in phaseC. the alternate antinodes vibrate in phaseD. all the particles between consecutive nodes vibrate in phase.

A standing wave is produced on a string on a string clamped at one end and free at the other. The length of the stringA. must be an integral multiple `lamda/4`B. must be an integral multiple of `lamda/2`C. must be an integral multiple of `lamda`D. may be an integral multiple of `lamda/2`

A wave is represented by the equation `y=(0.001mm)sin[(50s^-10t+(2.0m^-1)x]`A. The wave velocity `=100 ms^-1`B. The wavelength =2.0mC. the frequencey `=25/piHz`D. The amplitude `=0.001mm

A wave moving in a gasA. must be longitudinalB. may be longitudinalC. must be transverseD. may be transverse.

Two particles A and B have a phase diference of `pi` when a sine wave passes through the reginA. A oscillates at half the frequency of BB. A and B move in opposite directionsC. A and B must be separated by half of the wavelength.D. The displacements at A and B have equal magnitudes

Two blocks each having a mass of 3.2 kg are connected by a wire CD and the system is suspended from the ceiling by another wire AB . The linear mass density of the wire AB is 10 g m4 and that of CD is `8 g m^-1.` Find the speed of a transverse wave pulse produced in AB and in CD.

The length of the wire shown in figure between the pulley is 1.5 m and its mass is 12.0 g. Find the frequency of vibration with which the wire vibrates in two loops leaving the middle point of the wire between the pulleys at rest.

Which of the following equations represents as wave travelling along Y-axis?A. `x=Asin(ky-omegat)`B. `y=Asin(kx-omegat)`C. `y=Asinkycosomegat`D. `y=Acoskysinomegat`.

A sine wave is travelling in a medium. A particular partile has zero displacement at a certain instant. The particle closest to it having zero displacement is at a distanceA. `lamda/4`B. `lamda/3`C. `lamda/2`D. `lamda`

Velocity of sound in air is 332 `ms^-1`. Its velocity in vacuum will beA. `gt332ms^-1`B. `=332ms^-1`C. `lt332ms^-1`D. meaningless

Two waves passing through a region are represented by `y=(1.0cm) sin [(3.14 cm^(-1))x - (157s^(-1) x - (157s^(-1))t]` and `y = (1.5 cm) sin [(1.57 cm^(-1))x- (314 s^(-1))t].` Find the displacement of the particle at x = 4.5 cm at time t = 5.0 ms.