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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A `4.0 kg` block is suspended from the celling of an elevator through a string having a linear mass density of `1.6 xx 10^(-2) kg//m`. Find (a) speed (with respect to the string) with which pulse can proceed on the string it he elevator accelerates up at the rate of `6 m//s`. (b) in part (a) at some instant speed of lift is `40 m//s` in upward direction, then speed of the wave pulse with respect to ground at that instant is (Take `= 10 m//s^(2)`) |
| Answer» Correct Answer - (a) `200 m//s` , (b) `240 m//s` | |
| 2. |
A sonometer wire of length `1.5m` is made of steel. The tension in it produces an elastic strain of `1%`. What is the fundamental frequency of steel if density and elasticity of steel are `7.7 xx 10^(3) kg//m^(3)` and `2.2 xx 10^(11) N//m^(2)` respectively ?A. `188.5 Hz`B. `178.2 Hz`C. `200.5 Hz`D. `770 Hz` |
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Answer» Correct Answer - 2 `f = (v)/(2l) = (1)/(2l)sqrt((T)/(mu)) = (1)/(2l)sqrt((T)/(Ad))` Also `Y = (Tl)/(ADeltal) rArr (T)/(A) = (YDeltal)/(l) rArr f = (1)/(2l)sqrt((yDeltal)/(ld))` `l = 1.5m, (Deltal)/(l) = 0.01, d = 7.7 xx 10^(3) kg//m^(3)` After solving `f = sqrt((2)/(7)) xx (10^(3))/(3) Hz`. `f = 178.2 Hz`. |
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| 3. |
If the tension in a string is increased by `21` percent, the fundamental frequency of the string changes by `15 Hz`. Which of the following statements will also be correct?A. The original fundamental frequency is nearly `150 Hz`B. The velocity of propagation changes nearly by `4.5%`C. The velocity of propagation changes nearly by `10%`D. The fundamental wavelength changes nearly by `10%` |
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Answer» Correct Answer - A::C `V prop sqrt(T)` `f prop sqrt(T)` `(t)/(f + 15) = ((T)/(T + (21T)/(100)))^(1/2) = (10)/(11)` `11f = 10f + 150` `f = 150 Hz` |
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| 4. |
The same progressive wave is represented by two graphs I and II. Graph I shows how the displacement `y` varies with the distance x along the wave at a given time. Graph II shows how y varies with time t at a given point on the wave. The ratio of measurements AB to CD, marked on the curves represents:A. wave number `k`B. wave speed `V`.C. frequency `v`.D. angular frequency `omega`. |
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Answer» Correct Answer - B `("measure"AB)/("measure"CD) = (lambda)/(mu) = V` |
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| 5. |
A transverse periodic wave on a string with a linear mass density of 0.200 kg/m is described by the following equation `y=0.05 sin (420t-21.0 x)` where x and y are in metres and t is in seconds.The tension in the string is equal to :A. `32 N`B. `42 N`C. `66 N`D. `80 N` |
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Answer» Correct Answer - D `V_(omega) = (omega)/(k) = (420)/(21) = 20 :. V = sqrt((T)/(mu)) = 20 rArr T = (20)^(2)mu = 20^(2) xx 0.2 = 80N` |
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| 6. |
Derive an expression for the velocity of pulse in a in stretched in string under a tension `T` and `mu` is the mass per unit length of the sting. |
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Answer» Correct Answer - The broad view of the pulse under the tension `T` is given by As; `2T sin (theta)/(2) = dm (v^(2))/(R)` `rArr 2T(theta)/(2) = ((m)/(l)). Rtheta (v^2))/(R)` `rArr v = sqrt((T)/(mu))` where `mu = (m)/(l)`, is the mass per unit length of the string also known as the linear mass density. |
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| 7. |
Graph shows three waves that are separately sent along a string that is stretched under a certain tension along x-axis. If `omega_(1),omega_(2) and omega_(3)` are their angular frequencies, respectively, then: A. `omega_(1) = omega_(3) gt omega_(2)`B. `omega_(1) gt omega_(2) gt omega_(3)`C. `omega_(2) gt omega_(1) = omega_(3)`D. `omega_(1) = omega_(2) = omega_(3)` |
| Answer» Correct Answer - A | |
| 8. |
A piano string having a mass per unit length equal to `5.00 xx 10^(3)kg//m` is under a tension of `1350 N`, Find the speed with which a wave travels on this string. |
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Answer» Correct Answer - `300 sqrt(3) m//s` `V = sqrt((T)/(mu)) = ((1350)/(5 xx 10^(-3)))^(1/2) = 300sqrt(3) (m)//(sec)` |
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| 9. |
A guitar string is `180 cm` long and has a fundamental frequency of `90 Hz`. Where should it be pressed to produce a fundamental frequency of `135 Hz`? |
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Answer» Correct Answer - `120 cm` from on end. `f = (1)/(2L) sqrt((F)/(mu))` as `F` and `mu` are fixed `(f_(1))/(f_(2)) = (L_(2))/(L_(1))` `L_(2) = (f_(1))/(f_(2)) L_(1) = (90)/(135) xx 180cm = 120 cm` thus the string should be present at `120 cm` from an end |
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| 10. |
A piano wire weighting `4 g` and having a length of `90.0 cm` emits a fundamental frequency corresponding to the `"Middle C" (v = 125 Hz)`. Find the tension in the wire. |
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Answer» Correct Answer - `225 N` `l = 90 cm = 0.9m` `mu = (4)/(90) gm//cm` `= (4)/(900) Kg//m` `f = 125` `f = (1)/(2L) sqrt((T)/(mu))` `125 = (1)/(2 xx 0.9) sqrt((T xx 900)/(4))` `T = 225N` |
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| 11. |
A wave representing by the equation `y = a cos(kx - omegat)` is suerposed with another wave to form a stationary wave such that point `x = 0` is a node. The equation for the other wave isA. `a sin (kx + omegat)`B. `-a cos(kx + omegat)`C. `-a cos(kx - omegat)`D. `-a sin(kx - omegat)` |
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Answer» Correct Answer - B As `x = 0` is node `rArr` standing wave should be `y = 2a sin kx omegat` Now solve |
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| 12. |
What are the important effects which arises due to superposition of waves ? |
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Answer» Correct Answer - The superposition of two waves gives rise to the following important effects : 1. When two waves of same frequency (or wavelength) moving with the same speed and in the same direction superpose on each other, they give rise to to an effect, called interface of waves. 2. When two waves of the same frequency moving with the same speed in oppsite direction superpose on each other, they give rise to stationary waves. 3. When two waves of slightly different frequency moving with the same speed in the same direction superpose on each other, they give rise to beats. |
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| 13. |
A harmonic waves is travelling on string `1`. At a junction with string `2` it is partly reflected and partly transmitted. The linear mass density of the second string is four time that of the first string. And that the boundry between the two strings is at `x = 0`. If tje expression for the incident wave is, `y_(i) = A_(i) cos (k_(1)x - omega_(1)t)` What are the expression for the transmitted and the reflected waves in terms of `A_(i), k_(1)` and `omega_(1)` ? |
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Answer» Since `v = sqrt(T//mu), T_(2) = T_(1)` and `mu_(2) = 4mu_(1)` we have, `v_(2) = (v_(1))/(2)….(i)` The frequency does not changes, that is, `omega_(1) = omega_(2) …..(ii)` Also, because `k = omega//v`, the wave numbers of the harmonic waves in the two strings are releated by. `k_(2) = (omega_(2))/(v_(2)) = (omega_(1))/(v_(1)//2) = 2(omega_(1))/(omega_(1)) = 2k_(1) ....(iii)` The amplitudes are, `A_(t)=((2v_(2))/(v_(1)+v_(2)))A_(i)=[(2(v_(1)//2))/(v_(1)+(v_(1)//2))]A_(1)=(2)/(3)A_(1)...(iv)` and `A_(r)=((v_(2)-v_(1))/(v_(1)+v_(2)))A_(i)=[((v_(1//2))-v_(1))/(v_(1)+(v_(1)//2))]A_(i) = (A_(i))/(3)....(v)` Now with equation `(ii), (iii)` and `(iv)`, the transmitted wave can be written as, `y_(t) = (2)/(3)A_(i)cos(2k_(1)x - omega_(1)t)` Similarly the reflected wave can be expressed as, `= (A_(i))/(3)cos (k_(1)x + omega_(1)t + pi)` |
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| 14. |
Define superposition principal ? |
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Answer» Correct Answer - Superposition principle When two or more than two trains of waves travel in a medium at time each train of wave proceeds in dependently, as if the other trains of waves are in absent. The displacement of the particle at any instant of time due to these waves is totally independent of one another and can be obtained by the vector sum of the individuals displacements. Such a process by which the different trains of waves travelling through a medium simtultaneously overlap. one another without losing their individual nature or shape in called superposition of waves. |
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| 15. |
A travelling wave of amplitude `5 A` is partically reflected from a bounday with the amplitude `3A`. Due to superposition of two waves with different amplitude in opposite direction a standing wave pattern is formed. Datermine the ratio of amplitude at antinode to node. |
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Answer» Correct Answer - 4 `A_(i) = 5 A` `A_(r) = 3 A` As it is the carer of partially standing wave `:. A_(NODE) = A_(i) - A_(r)` `= 2A` and `A_(ANTINODE) = A_(i) + A_(r) = 8A` |
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| 16. |
Sinusoidal waves 5.00 cm in amplitude are to be transmitted along a string having a linear mass density equal to `4.00xx10^-2 kg//m`.If the source can deliver a average power of 90 W and the string is under a tension of 100 N, then the highest frequency at which the source can operate is (take `pi^2=10`):A. `45 Hz`B. `50 Hz`C. `30 Hz`D. `62 Hz` |
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Answer» Correct Answer - C As `ltPgt = 2pi^(2)f^(2)A^(2)muv` put values `90 = 2 xx 10 xx f^(2) xx 25 xx 10^(-4) xx 4 xx 10^(-2)sqrt((100)/(4 xx 10^(-2))) rArr = f = 30 Hz` |
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| 17. |
A series of pulses, each of amplitude `0.150 m`, are sent on a string that is attached to a wall at one end. The pulses are reflected at the wall and travel back along the string without loss of amplitude. When two waves are present on the same string. The net displacement of a given point is the sum of the displacement of the individuals waves at the point. What is the net displacement at point on the spring where two pulses am crossing, (a) if the string is rigidly attached to the post? (b) if the end at which reflection occurs is free is slide up and down ? |
| Answer» Correct Answer - (a) Zero , (b) `0.300 m`. | |
| 18. |
Two wave pulses travel in opposite directions on a string and approach each other. The shape of one pulse is inverted with respect to the otherA. the pulses will collide with each other and vanish after collisionB. the pulses will reflect from each other i.e., the pulse going towards right will finally move towards left and vice versa.C. the pulse will pass through each other but their shapes will be modifiedD. the pulse will pass through each other without any change in their shape |
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Answer» Correct Answer - D By difination |
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| 19. |
A transverse sinusoidal wave is generted at one end of long, horizontal string by a bar that moves up and down through a distance of `1.00 cm`. The motion is continuous and is repreated regularly `120` times per second. The string has linear density `90 gm//m` and is kept under a tension of `900 N`. Find : What is the maximum power (in watt) transferred along the string.A. `3.24 pi^(2)`B. `6.48 pi^(2)`C. `12.96 pi^(2)`D. `25.92 pi^(2)` |
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Answer» Correct Answer - C `P_(max) = 4pi^(2)f^(2)A^(2)muv = 12.96pi^(2)` watt |
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| 20. |
A transverse sinusoidal wave is generted at one end of long, horizontal string by a bar that moves up and down through a distance of `1.00 cm`. The motion is continuous and is repreated regularly `120` times per second. The string has linear density `90 gm//m` and is kept under a tension of `900 N`. Find : The maximum value of the transverse component of the tension (in newton)A. `1.8 pi`B. `1.08 pi`C. `9`D. `18 pi` |
| Answer» Correct Answer - B | |
| 21. |
A wave pulse is generated in a string that lies along `x-`axis. At the points `A` and `B`, as shown in figure, if `R_(A)` and `R_(B)` are ratio of magnitudes of wave speed to the particle speed, then A. `R_(A) gt R_(B)`B. `R_(B) gt R_(A)`C. `R_(A) = R_(B)`D. Information is not sufficient to decide. |
| Answer» Correct Answer - B | |
| 22. |
A transverse sinusoidal wave moves along a string in the positive x-direction at a speed of `10m//s`. The wavelength of the wave is `0.5m` and its amplitude is `10cm`. At a particular time `t`, the snap-shot of the wave is shown in figure. The velocity of point `P` when its displacement is `5cm` is - A. `(sqrt(3)pi)/(50)hat(j) m//s`B. `-(sqrt(3)pi)/(50)hat(j) m//s`C. `(sqrt(3)pi)/(50)hat(i) m//s`D. `-(sqrt(3)pi)/(50)hat(i) m//s` |
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Answer» Correct Answer - 5 `A_(eq) = sqrt(A_(1)^(2) + A_(2)^(2) + 2A_(1)A_(2) cos phi)` `A_(eq) = sqrt(4^(2) + 3^(2) + 2(4)(3)cos"(pi)/(2))` `A_(eq) = 5`. |
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| 23. |
A taut string having tension 100 N and linear mass density `0.25 kg//m` is used inside a cart to generate a wave pulse starting at the left end, as shown. What should be the velocity of the cart so that pulse remains stationary w.r.t. ground. |
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Answer» Velocity of pulse `= sqrt((T)/(mu)) = 20 m//s` Now `overset(vec)(v)_(PG) = overset(vec)(v)_(PC) + overset(vec)(v)_(CG)` `0 = 20 I + overset(vec)(v)_(CG)` `overset(vec)(v)_(CG) = -20i m//s` |
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| 24. |
A way pulse is travelling on a string of linear mass density `6.4 xx 10^(-3)kg m^(-1)` under a load of `80 kgf`. Calculate the time taken by the pulse to traverse the string, if its length is `0.7 m`. |
| Answer» Correct Answer - `2 xx 10^(-3)s` | |
| 25. |
A transverse sinusoidal wave of amplitude `a`, wavelength `lambda` and frequency `f` is travelling on a stretched string. The maximum speed of any point in the string is `v//10`, where `v` is the speed of propagation of the wave. If `a = 10^(-3)m` and `v = 10ms^(-1)`, then `lambda` and `f` are given byA. `lambda = 2pi xx 10^(-2)m`B. `lambda = 10^(-2)cm`C. `f = 10^(3)/(2pi)Hz`D. `f = 10^(4)Hz` |
| Answer» Correct Answer - A::C | |
| 26. |
You have learnt that a travelling wave in a dimension is represented by a funcation `y = f(x,t)` where `x` and `t` must appear in the combination `x-vt` or `x + vt`, i.e `y = f(x+- upsilon t)`. Is the coverse true? (Examine if the following funcations for `y` can possibly represent a travelling wave : (a) `(x - vt)^(2)` (b) `log [(x + v)//x_(0)]` (c) `1//(x + v)` |
| Answer» Correct Answer - The converse is not true. An obvious requirement for an acceptable function for a travelling wave is that it should be finite everywhere and at all times. Only funcation © satisfies this condition, the remaining funcations cannot possibly represent a travelling wave. | |
| 27. |
The wave funcation for a travelling wave on a string in given as `y (x, t) = (0.350 m) sin (10 pi t - 3pix + (pi)/(4))` (a) What are the speed and direction of travel of the wave ? (b) What is the vertical displacement of the string at `t = 0, x = 0.1 m` ? (c) What are wavelength and frequency of the wave ? |
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Answer» `Y(x,t) = (0.350m) sin (10pit - 3pi x + (pi)/(4))` comparing with equation , `Y = A sin (omegat - kx + phi) , omega = 10 pi , k = 3pi, f = (pi)/(4)` (a) speed `= (omega)/(k) = (10)/(3) = 3.33 m//sec` and along `+ve x` axis (b) `y(0.1, 0) = 0.35 sin (10 pi xx O - 3 pi (0.1) + (pi)/(4)) = 0.35 sin [(pi)/(4) - (3pi)/(10)] = -5.48 cm` (c) `k = (2pi)/(lambda) = 3 pi rArr lambda = (2)/(3) cm , = 0.67 cm` and `f = (v)/(lambda) = (10//3)/(2//3) = 5 Hz`. |
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| 28. |
For the wave shown in figure, the equation for the wave, travelling along `+x` acis with velocity `350 ms^(-1)` when its position at `t = 0` is as shown A. `0.05 sin ((314)/(4)x - 27475 t)`B. `0.05 sin ((379)/(5)x - 27475 t)`C. `1 sin ((314)/(4)x - 27475 t)`D. `0.05 sin ((289)/(5)x + 27475 t)` |
| Answer» Correct Answer - A | |
| 29. |
A wire having a linear mass density `10^(-3) kg//m` is stretched between two rigid supports with a tension of `90 N`. This wire resontes at a frequency of `350 Hz`. The next higher frequency at which the same wire resonates is `420 Hz`. Find the length of the wire. |
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Answer» Suppose the wire vibrates at `350 Hz` in its nth harmonic and at `420 Hz` in its `(n + 1)` th harmonic. `350s^(-1) = (n)/(2 L) sqrt((F)/(mu)) ….(i)` and `420s^(-1) = ((n + 1))/(2l.) sqrt((F)/(mu)) …..(ii)` This gives `(420)/(350) = (n + 1)/(n)` or, `n = 5`. Putting the value in `(i)` `350 = (5)/(2l)sqrt((90)/(10^(-3))) rArr 350 = (5)/(2l) xx 300 rArr l = (1500)/(700) = (15)/(7)m = 2.1 m` |
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| 30. |
A circular loop of rope of length L rotates with uniform angular velocity `omega` about an axis through its centre on a horizontal smooth platform. Velocity of pulse (with resppect to rope ) produce due to slight radiul displacement is given by A. `omegaL`B. `(omegaL)/(2pi)`C. `(omegaL)/(pi)`D. `(omegaL)/(4pi^(2))` |
| Answer» Correct Answer - B | |
| 31. |
A wire of `9.8 xx 10^(-3) kg` per meter mass passes over a fricationless pulley fixed on the top of an inclined frictionless plane which makes an angle of `30^(@)` with the horizontal. Masses `M_(1)` & `M_(2)` are tied at the two ends of the wire. The mass `M_(1)` rests on the plane and the mass `M_(2)` hangs freely vertically downwards. the whole system is in equilibrium. Now |
| Answer» Correct Answer - 2 | |
| 32. |
A non-uniform rope of mass `M` and length `L` has a variable linear mass density given by `mu = kx`, where `x` is the distance from one end of the wire and `k` is constant. `t = sqrt((pML//9F))` where `F` (constant) is the tension in the wire then and `p` |
| Answer» Correct Answer - 8 | |
| 33. |
Three blocks `I, II`, & `III` having mass of `1.6 kg`, `1.6 kg` and `3.2 kg` repectively are connected as shown in the figure. The linear mass density of the wire `AB, CD` and `DE` are `10 g//m, 8 g//m` and `10 g//m` respectively. The speed of a transverse wave pulse produced in `AB, CD` and `DE` are : `(g = 10m//sec^(2))` A. `80 m//s, 20sqrt(10) m//s, 40 m//s`B. `20 sqrt(10) m//s, 40 m//s, 80 m//s`C. `20 sqrt(10) m//s` in allD. `80 m//s` in all |
| Answer» Correct Answer - A | |
| 34. |
Both the strings, shown in figure, are made of same material and have same cross-section. The pulleys are light. The wave speed of transverse wave in the string `AB` is `v_(1)` and in `CD` it is `v_(2)`, the `v_(1)//v_(2)` is A. `1`B. `2`C. `sqrt(2)`D. `(1)/(sqrt(2))` |
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Answer» Correct Answer - C `V prop sqrt(1)` `(V_(1))/(V_(2)) = sqrt((T_(1))/(T_(2))) = sqrt((2T)/(T)) = sqrt(2)` |
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| 35. |
A nylon guiter string has a linear density of `7.20 g//m` and is under a tension of `150 N`. The fixed supports are distance `D = 90.0 cm` apart. The string is osillating in the stading wave pattern shown in figure. Calculate the (a) speed. (b) wavelength and (c) frequency of the travelling waves whose superposition gives this standing wave. ltbgt |
| Answer» Correct Answer - (a) `(250)/(sqrt(3))m//s;` , (b) `60.0 cm;` , (c) `(1250)/(3sqrt(3)) Hz` | |
| 36. |
The particles displacement in a wave is given by `y = 0.2 xx 10^(-5) cos (500 t - 0.025 x)` where the distances are measured in meters and time in seconds. NowA. wave velocity is `2 xx 10^(4)ms^(-1)`B. particles velocity is `2 xx 10^(4)ms^(-1)`C. initial phase difference is `(pi)/(2)`D. wavelength of the wave is `(80pi)m` |
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Answer» Correct Answer - A::D Comparing with `y = A cos (omegat - kx)` `omega = 500s^(-1), k = 0.025 m^(-1)`, `v = (500)/(0.025) = 2 xx 10^(4)m//s , lambda = (2pi)/(0.025) = 80 pi m` `y = A cos (omegat - kx)` |
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| 37. |
A wave equation which gives the displacement along the y direction is given by `y=10^(-4)sin(60t+2x)`, where x and y are in meters and t is time in seconds This represents a waveA. travelling with a velocity of `30 m//s` in the negative `x` directionB. of wavelength `pi` meterC. of frequency `30//pi` hertzD. of amplitude `10^(-4)` meter travelling along the negative `x` direction. |
| Answer» Correct Answer - A::B::C::D | |
| 38. |
Two waves passing through a region are respresented by `y_(1) = 5 mm sin [(2pi cm^(-1))x - (50 pis^(-1))t]` and `y_(2) = 10 mm sin [(pi cm^(-1))x - (100 pis^(-1))t]` Find the displacement of the particle at `x = 1 cm` at time `t = 5.0 ms`. |
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Answer» Accoding to the principle of superposition, each wave produce its disturbance independent of the other and the resultant disturbance is equal to the vector sum of the individual disturbance. The displacements of the particle at `x = 1cm` at time `t = 5.0 ms` due to the two waves are. `y_(1) = 2 mm [(2pi cm^(-1)) xx - (50 pi s^(-1))t]` `y_(1) = 5 mm sin[(2pi cm^(-1)) xx 1 cm - (50 pi s^(-1))5 xx 10^(-3) sec]` `= 5 mm sin [2pi - (p)/(4)] = -5 mm` and `y_(2) = 10 mm sin [(pi cm^(-1)) xx - (100 pi s^(-1))t]`, `y_(2) = 10 mm sin [(pi cm^(-1)) xx 1 cm - (100 pi s^(-1))5 xx 10^(-3) sec]` `= 10 mm sin [pi - (pi)/(2)] = 10 mm` The net displacement is : `y = y_(1) + y_(2) = 10 mm - 5 mm = 5 mm` |
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| 39. |
The vibration of a string of length `60 cm` is represented by the equation, `y = 3 cos (pix//20) cos(72pit)` where `x` & `y` are in `cm` and `t` in sec. (i) Write down the compoent waves whose superposition gives the above wave. (ii) Where are the nodes and antinodes located along the string. (iii) What is the velocity of the particle of the string at the position `x = 5 cm` & `t = 0.25` sec. |
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Answer» Correct Answer - (i) `y_(1) = 1.5 cos {(pi//20)x - 72pit}`, `y_(2) = 1.5 cos {(pi//20)x + 72 pit)` (ii) `10, 30, 50 cm` and `0 , 20 , 40 , 60 cm` (iii) `0` |
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| 40. |
A transverse wave travelling in a string produce maximum transverse velocity of `3 m//s` and maximum transverse acceleration `90 m//s^(2)` in a particle. If the velocity of wave in the string is `20 m//s`. Datermine the equation of the wave ? |
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Answer» Correct Answer - Equation of wave in string `y = 0.1 sin (30t +- (3)/(2) x + phi)` [where `phi` is initial phase] Maximum particle velocity `v_(max) = omegaA = 3m//s` Maximum particle accelertion `a_(max) = omega^(2) A = 90m//s^(2)` `a_(max) = v_(max) omega = omega xx 3 = 90 m//s^(2) rArr omega = 30s^(-1) rArr A = (3)/(omega) = (3)/(30) = 0.1 m` `k = (omega)/(v) = (30)/(20) = (3)/(2)` [ where `v` is velocity of wave] Equation of wave in string `y = 0.1 sin (30t+-(3)/(2)xrarr)` where `phi` is initial phase] |
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| 41. |
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 [2pi((t)/(0.04(s))-(x)/(0.50(m)))]`. The tension in the string is :A. `4.0 N`B. `12.5 N`C. `0.5 N`D. `6.25 N` |
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Answer» Correct Answer - 4 By equation `f = (1)/(0.04)` and `lambda = 0.5` `rArr V = (1)/(0.04) xx 0.5 = (25)/(2)` by `V = sqrt((T)/(mu)) rArr ((25)/(2))^(2) = (T)/(0.04) rArr T = (625)/(4) xx 0.04` `T = 6.25 N` |
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| 42. |
Find velocity of wave is string `A` &`B`. |
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Answer» Correct Answer - `50 m//sec, 25 m//sec` `V_(1) = sqrt((10)/(4 xx 10^(-3))` `= 50 m//sec` `V_(2) = sqrt((10)/(16 xx 10^(-3)))` `= sqrt((10^(4))/(16)) = (100)/(4) = 25 m// sec`. |
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| 43. |
A heavy ball is suspended from the ceilling of a motor can through a light strig. A transverse pulse travels at a speed of `50 cm//s` on the strin when the car is at rest and `52 cm//s` when the car accelerates on a horizontal road. Then acceleration of the car is : (Take `g = 10 m//s^(2)`.)A. `2.7 m//s^(2)`B. `3.7 m//s^(2)`C. `2.4 m//s^(2)`D. `4.1 m//s^(2)` |
| Answer» Correct Answer - D | |
| 44. |
Consider the wave `y = (10 mm) sin[(5picm^(-1))x-(60pis^(-1))t + (pi)/(4)]`. Find (a) the amplitude (b) the wave number (c) the wavelength (d) the frequency f the time period and (f) the wave velcity (g) phase constant of `SHM` of particle at `x = 0`. |
| Answer» Correct Answer - (a) `10 mm` , (b) `5 pi cm^(-1)` , (c) `30 Hz` , (d) `30 Hz` , (e) `(1)/(30)s` , (f) `12 cm//s` | |
| 45. |
In an experiment of standing waves, a string `90 cm` long is attached to the prong of an electrically driven tuning fork that scillates perpendicular to the length of the string at a frequency of `60 Hz`. The mass of the string is `0.044 kg`. What tension (in newton) must the string be under (weights are attached to the other end) if it is to oscillate in four loops? |
| Answer» Correct Answer - 36 | |