32 + Interview Questions in SUPERPOSITION-OF-WAVES IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Among two interfering sources, let `S_1` be ahead of the phase by `90^@` relative to `S_2` . If an observation point P is such that `PS_1 - PS_2 = 1.5 lambda` , the phase difference between the waves from `S_1 and S_2` reaching P isA. `3pi`B. `(5pi)/2`C. `(7pi)/2`D. `4pi`
Answer» Correct Answer - B `Delta phi_1 = pi/2` `Delta phi_2 = ((2pi)/lambda) (Deltax)` `= ((2pi)/lambda) (1.5 lambda)` ` = 3pi ` `:. Delta phi_("net") = Delta phi_2 - Delta phi_1 = (5pi)/2` .

Discussion

2.	A string that is 10 cm long is fixed at both ends. At t=0, a pulse travelling from left to right at `1cm//s is 4.0 cm ` from the right end as shown in figure. Determine the next two times when the pulse will be at that point again. State in each case whether the pulse is upright or inverted . .
Answer» Correct Answer - B::D `t_(1) = d_(1)/v = (4+4)/1 = 8s`. After one reflection from fixed boundary wave is inverted. `t_2 = d_(2)/v = (4+10+16)/1 = 20s ` After two times reflection from fixed boundaries wave pulse will again become upright.

Discussion

3.	A man generates a symmetrical pulse in a string by moving his hand up and down. At `t=0` the point in his hand moves downward. The pulse travels with speed of `3 m//s` on the string & his hands passes `6` times in each second from the mean position. Then the point on the string at a distance `3m` will reach its upper extreme first time at time `t=`A. 1.25 sB. 1sC. `11//12 s `D. 23/24 s
Answer» Correct Answer - C `f = 3Hz` `:. Omega = 2pif = (6pi) "rad"//s ` `k = omega/v = (6pi)/3 = (2pi) "rad"//m` ` y = A sin (kx-omegat)` ` = A sin (2pix - 6pit)` Putting `y= +-A` and `x = 3`, we get ` (2pi)(3) - 6pi t = pi/2 ` `:. 6pit = 11/2 pi ` ` :. t = 11/12s `.

Discussion

4.	A standing wave is formed by two harmonic waves, `y_1 = A sin (kx-omegat) and y_2 = A sin (kx + omegat)` travelling on a string in opposite directions. Mass density energy between two adjavent nodes on the string.
Answer» Correct Answer - A::B The distance between two adjacent nodes is ` lambda/2 or pi/k.` `:. Volume of string between two nodes will be ` `V = ("area of cross-section") xx ("distance between two nodes")` `=(S) (pi/k)` Energy density (energy per unit volume) of a travelling wave is given by `u=1/2 rho A^2 omega^2` A standing wave is formed by two identical waves travelling in opposite directions. Therefore, the energy stored between two nodes in a standing wave. `E= 2`[energy stored in a distance of `pi/k` of a tracelling wave] = 2 (energy density)(volume) = 2 (1/2 rho A^2 omega^2) ((piS)/k) or `E = (rho A^2 omega^2 piS)/k`

Discussion

5.	`y_1 = 8 sin (omegat - kx) and y_2 = 6 sin (omegat + kx)` are two waves travelling in a string of area of cross-section s and density rho. These two waves are superimposed to produce a standing wave. (a) Find the energy of the standing wave between two consecutive nodes. (b) Find the total amount of energy crossing through a node per second.
Answer» Correct Answer - A::B::C (a) Distance between two nodes is `lambda/2` and `pi/k`. The volume of string between two nodes is `V = pi/k s` ......(i) Energy density (energy per unit volume ) of each wave will be `u_1 = 1/2 rho omega^2 (8)^2 = 32 rho omega^2` and ` u_2 = 1/2rho omega^2 (6)^2 = 18 rho omega^2 ` `:.` Total mechanical energy between two consecutive nodes will be `E = (u_1 + u_2)V` `= 50 pi/k rho omega^2S` (b) `y= y_1 + y_2 ` `= 8 sin (omegat - kx) + 6 sin (omegat + kx)` ` = 2 sin (omegat - kx) + { 6 sin (omegat + kx)` ` + 6sin (omegat + kx)}` `= 2 sin (omegat - kx ) + 12 cos kxsin omegat ` Thus, the resultant wave will be a sum of standing wave and a travelling wave. Energy crossing through a node per second = power or travelling wave. `:. P = 1/2 rho omega^2 (2)^2 Sv ` `= 1/2 rho omega ^2 (4)(S)(omega/k)` `= (2rho omega^3 S)/k` .

Discussion

6.	A wave travels on a light string. The equation of the waves is `y= A sin (kx - omegat + 30^@)`. It is reflected from a heavy string tied to an end of the light string at x = 0. If 64% of the incident energy is reflected, then the equation of the reflected wave isA. `y = 0.8 A sin (kx - omegat + 30^@ + 180^@)`B. `y = 0.8 A sin (kx + omegat + 30^@ + 180^@)`C. `y= 0.8 A sin (kx - omegat - 30^@)`D. `y= 0.8 A sin (kx - omegat + 30^@)`
Answer» Correct Answer - B `I prop A^2` ` I_r = 0.64 I_i` ` :. A_r = 0.8 A_i` ` = 0.8 A ` Reflected from a denser medium. Hence, a phase change of `pi` will occur. Reflected wave will travel in opposite direction. Hence, `omegat and kx` will have positive signs.

Discussion

7.	Which of the following equations can form stationary waves? (i)`y= A sin (omegat - kx)` (ii) `y= A cos (omegat - kx)` (iii) `y= A sin (omegat + kx)` (iv) `y= A cos (omegat - kx)` .A. (i) and (ii)B. (i) and (iii)C. (iii) and (iv)D. (ii) and (iv)
Answer» Correct Answer - B::D Two identical waves should travel in opposite directions.

Discussion

8.	Regarding stationary waves, choose the correct options.A. This is an example of interferenceB. Amplitudes of waves may be differentC. particles at nodes are always at restD. Energy is conserved
Answer» Correct Answer - A::B::D

Discussion

9.	Incident wave `y= A sin (ax + bt+ pi/2)` is reflected by an obstacle at x = 0 which reduces intensity of reflected wave by 36%. Due to superposition, the resulting wave consists of a standing wave and a travelling wave given by `y= -1.6 sin ax sin bt + cA cos (bt + ax)` where A, a, b and c are positive constants. 3. Position of second antinode isA. `x = pi/(3a)`B. `x = (3pi)/a`C. `x = (3pi)/(2a)`D. `x = (2pi)/(3a)`
Answer» Correct Answer - C `A_x = -1.6 sin ax ` `:. X = 0` is a node ` ` Second antinode is at a distance. ` `x = lambda/4 + lambda/2 = (3lambda)/4 = 3/4 (2pi)/k` But, k=a `:. x = (3pi)/(2a)`.

Discussion

10.	Assertion: Stationary waves are so called because particles are at rest in stationary waves. Reason: They are formed by the superposition of two identical waves travelling in opposite directions.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, but the Reason is false.D. If Assertion is false but the Reason is true.
Answer» Correct Answer - D They are called stationary because net energy transfers from any section is zero, if amplitudes of constituent waves are zero.

Discussion

11.	Incident wave `y= A sin (ax + bt+ pi/2)` is reflected by an obstacle at x = 0 which reduces intensity of reflected wave by 36%. Due to superposition, the resulting wave consists of a standing wave and a travelling wave given by `y= -1.6 sin ax sin bt + cA cos (bt + ax)` where A, a, b and c are positive constants. 2. Value of c isA. 0.2B. 0.4C. 0.6D. 0.3
Answer» Correct Answer - A `y = y_i + y_r ` ` = A sin (ax + bt + pi//2)` `+0.8 A sin (ax - bt +pi//2)` ` = [0.8 A sin (ax +bt + pi/2)` `+ 0.8 A sin (ax -bt + pi/2)] ` `+ 0.2 A sin (ax + bt+ pi/2)` `= 1.6 sin ax sin bt + 0.2 A cos (bt + ax)` `:. c = 0.2`

Discussion

12.	Two wires of different densities are soldered together end to end then stretched under tension T. The waves speed in the first wire is twice that in the second wire. (a) If the amplitude of incident wave is A, what are amplitudes of reflected and transmitted waves? (b) Assuming no energy loss in the wire, find the fraction of the incident power that is reflected at the junction and fraction of the same that is transmitted.
Answer» Correct Answer - A::B::C `v_2 = (v_1)/2` (a) ` A_r = ((v_2-v_1)/(v_1+v_2))A` `=[((v_1//2)-v_1)/(v_1+(v_1//2))]A` ` = -A/3` `A_t = ((2v_2)/(v_1+v_2))A` `=[(2xx((v_1)/2))/(v_1+((v_1)/2)] A` `= 2/3 A` . (b) Since, reflected waves come in the same medium, we can say that `P prop A^2` `P_r/P_i = (A_r/A_i)^2 = ((-A//3)/A)^2 = 1/9` :. Fraction of power transmitted `=1-1/9 = 8/9`.

Discussion

13.	The tension, length, diameter and density of a string B are double than that of another string A. Which of the following overtones of B is same as the fundamental frequency of A?A. 1stB. 2ndC. 3rdD. 4th
Answer» Correct Answer - C `eta (v/(2l))_B = (v/(2l))_A` `eta = sqrt((T//rhoS))/(2l))_B = (v/(2l))_A` or ` eta ((sqrt (4T//rhopi))/(2ld))_B = ((sqrt(4T//rhopi))/ (2ld))_A` ……….(i) Given, `T_B = 2T_A, l_B = 2l_A` d_B = 2d_A` and `rho_B = 2 rho_A` Putting in Eq. (i), we get `eta = 4 ` `eta = 4` means 4th harmonic or 3rd overtone.

Discussion

14.	When tension of a string is increased by 2.5 N, the initial frequency is altered in the ratio of 3:2. The initial tension in the string isA. 6NB. 5NC. 4ND. 2N
Answer» Correct Answer - D `:. f prop sqrtT` `:. f_1/f_2 = sqrt(T_1/T_2)` `:. 3/2 = sqrt((T+2.5)/T)` Solving, we get `T = 2N`.

Discussion

15.	The lengths of two wires of same material are in the ratio 1:2, their tensions are in the ratio 1:2 and their diameters are in the ratio 1:3. the ratio of the notes they emits when sounded together by the same source isA. `sqrt(2)`B. `sqrt3`C. `2sqrt(3)`D. `3sqrt(2)`
Answer» Correct Answer - D `f = v/(2l) = sqrt(T//mu)/(2l) = sqrt(T//rhoS)/l` `= sqrt(T//rho(pid^2//4))/l` or `f prop sqrtT/(dl)` `:. f_1/f_2 = (sqrt(T_1/T_2)) (d_2/d_1)(l_2/l_1)` `= sqrt 1/2 (3/1)(2/1) = 3sqrt(2)`.

Discussion

16.	A string 1m long is drawn by a 300 Hz vibrator attached to its end. The string vibrates in three segments. The speed of transverse waves in the string is equal toA. `100 m//s`B. `200 m//s`C. `300 m//s`D. `400 m//s`
Answer» Correct Answer - B `3(v/(2l)) = 300` `:. v=200 l=(200)(1)` `=200 m//s`.

Discussion

17.	Sound waves of frequency `600 H_(Z)` fall normally on perfectly reflecting wall. The distance from the wall at which the air particles have the maximum amplitude of vibration is (speed of sound in air = `330 m//s`)
Answer» Correct Answer - A::B Wall will be a node (displacement). Therefore, shortest distance from the wall at which air particles have maximum amplitude of vibration (displacement antinode) should be `lambda/4`. Here, `lambda = v/f = 330/660 = 0.5 m ` `:. Desired distance is `0.5/4 = 0.125 m`.

Discussion

18.	The ratio of intensities of two waves is 9:16. If these two waves interfere, then determine the ratio of the maximum and minimum possible intensities.
Answer» Correct Answer - A::D `I_(max)/I_(min) = ((sqrt((I_1//I_2))+1)/(sqrt((I_1//I_2))-1))^(2)`

Discussion

19.	Two identical harmonic pulses travelling in opposite directions in a taut string approach each other. At the instant when they completely overlap, the total energy of the string will be .A. zeroB. partly kinetic and partly potentialC. purely kineticD. purely potential
Answer» Correct Answer - C

Discussion

20.	A stationary wave is given by `y = 5 sin (pi x)/( 3) cos 40 pi t` where `x and y` are in `cm and t` is in second. a. What are the amplitude and velocity of the component wave whose superposition can give rise to this vibration ? b. What is the distance between the nodes ? c. What is the velocity of a particle of the string at the position ` x = 1.5 cm when t = 9//8 s`?
Answer» Correct Answer - A::B::C (a) `v= omega/k = (40pi)/(pi//3)` ` = 120 cm//s` (b) Distance between adjacent nodes ` = lambda/2 = pi/k = (pi/(pi//3))` = 3 cm (c) v_P = (dely)/(delt) = (-200pi) sin (pix)/3 sin 40pit` Now, substitute the given value of x and t.

Discussion

21.	The period of oscillations of a point is 0.04 s and the velocity of propagation of oscillation is 300 `m//s`. The difference of phases between the oscillations of two points at distance 10 m and 16 m respectively from the source of oscillations isA. `2pi`B. `pi/2`C. `pi/4`D. `pi`
Answer» Correct Answer - D `:. Deltaphi = ((2pi)/lambda) (Deltax)` `=((2pi)/(v//f))(Deltax)= ((2pi)/(vT))(Deltax)` `= ((2pi)/(300 xx 0.04)) (16-10)` `=pi`.

Discussion

22.	A wire fixed at both ends is 1.0 m long and has a mass of 36g. One of its oscillation frequencies is `250 Hz` and the next higher one is 300 Hz. (a) Which harmonics do these frequencies represent? (b) What is the tension in the wire?
Answer» Correct Answer - A::B::C::D (a) `f_1/f_2 = 250/300 = 5/6` So, `f_1` is `5th` harmonic and `f_2` is `6th` harmonic. (b) `f_1 = (5) = v/(2l) = (2.5 (sqrt(T/mu)))/(l)` `:. T = (mu f_(1)^(2)l^(2))/(2.5)^2) = ((0.036) (250)^2 (1)^2)/(2.5)^(2)` =360 N.

Discussion

23.	Two different stretched wires have same tension and mass per unit length. Fifth overtone frequency of the first wire is equal to second harmonic frequency of the second wire. Find the ratio of their lengths.
Answer» Correct Answer - A::B::C `6[(T//mu)/(2l_1)] = 2[(T//mu)/(2l_2)]` `:. l_(1)/l_(2) = 3 ` .

Discussion

24.	Find the fundamental frequency and the next three frequencies that could cause a standing wave pattern on a string that is 30.0 m long, has a mass per unit length of `9.00 xx 10^(-3) kg//m` and is stretched to a tension of 20.0 N.
Answer» Correct Answer - A::B::C::D `v= sqrtT/mu = sqrt (20/(9xx (10^-3))) = 47.14 m//s` `f_1 = v/(2l) = (47.14)/(2xx 30)` = 0.786 Hz. Next three frequencies are `2f_1, 3f_1 and 4f_1` .

Discussion

25.	In terms of T, `mu` and l, find frequency of (a) fourth overtone mode (b) third harmonic .
Answer» Correct Answer - A::B::C (a) Fourth overtone mode means `n=5 (n=1,2,3,4,5)` Therefore, using the equation `f=(n)/(21) sqrt((T)/(mu)) = (5)/(2l) sqrt((T)/(mu))` (b) Third harmonic means n=3 ` :. f=(3)/(2l) sqrt(T/mu)`.

Discussion

26.	If the frequencies of the sound and fifth harmonics of a string differ by 54 Hz. What is the fundamental frequency of the string ?
Answer» Correct Answer - A 5f - 2f = 54 `:. 3f = 54` `:. f = 18 Hz` .

Discussion

27.	Three resonant frequencies of a string are 90, 150 and 210 Hz. (a) Find the highest possible fundamental frequency of vibration of this string. (b) Which harmonics of the fundamental are the given frequencies ? (c) Which overtones are these frequencies ? (d) If the length of the string is 80 cm, what would be the speed of a transverse wave on this string ?A. `45 m//s`B. `75 m//s`C. `48 m//s`D. `80 m//s`
Answer» Correct Answer - C These are multiple of 30 Hz. Hence, fundamental frequency `f_(0) = 30Hz` Now, `f_(0) = v/(2l)` `:. v=2f_(0)l ` `=2xx 30 xx 0.8` `=48 m//s` .

Discussion

28.	For a certain stretched string, three consecutive resonance frequencies are observed as 105, 175 and 245 Hz respectively. Then, the fundamental frequency isA. 30 HzB. 45 HzC. 35 HzD. None of these
Answer» Correct Answer - C All frequencies are integral multiples of `35 Hz`.

Discussion

29.	A wire with mass 40.0 g is stretched so that its ends are tied down at points 80.0 cm apart.The wire vibrates in its fundamental mode with frequency 60.0 Hz and with an amplitude at the antinodes of 0.300 cm. (a) What is the speed of propagation of transverse wave in the wire? Compute the tension in the wire.(c) Find the maximum transverse velocity and acceleration of particles in the wire.
Answer» Correct Answer - A::B::C::D (a)`lambda/2 = l` or `lambda= 2l = 1.6 m ` `v = f lambda = 60 xx 1.6 = 96 m//s` `(b) v = sqrt(T/mu)` `:. T = muv^2 = (0.04/0.8) (96)^2` = 461 N (c) `v_(max) = omega A_(max)` ` = (2pif) A_(max)` ` = (2pi)(60)(0.003)` ` = 1.13 m//s ` ` a_(max) = omega^2 A_(max)` ` = (2pif)^2 A_(max)` `=(2pi xx 60)^2 (0.003)` `=426.4 m//s^2` .

Discussion

30.	If the tension in a stretched string fixed at both ends is changed by `21%` , the fundamental frequency is found to increase by `15 Hz`, then theA. original frequency is 150 HzB. velocity of propagation of the transverse wave along the string increases by 5%.C. velocity of propagation of the transverse wave along the string increases by 10% .D. fundamental wavelength on the string does not change.
Answer» Correct Answer - A::C::D `f prop sqrt T` `:. f_2/f_1 = sqrt (T_2/T_1)` `:. (f_1+15)/f_1 = sqrt(1.21T_1)/T_1` = 1.1 Solving we get `f_1 = 150 Hz` `v prop sqrtT` `:. v_2/v_1 = sqrt (T_2/T_1)` or ` v_2 = sqrt((1.21 T_1)/(T_1)) v_1 ` ` = 1.1 v_1` Hence, increase in v is 10% `lambda/2 =l ` `:. lambda = 2l ` `:. Fundamental wavelength = 2 lambda is unchanged. `

Discussion

31.	The harmonic wave `y_i = (2.0 xx 10^(-3)) cos pi (2.0x - 50t)` travels along a string towards a boundary at x=0 with a second string. The wave speed on the second string is `50 m//s`. Write expressions for reflected and transmitted waves. Assume SI units.
Answer» Correct Answer - A::B::C Speed of wave in first medium is, `v_1` = (Coefficient of `t`)/(Coefficient of `x`) = `50/2 = 25 m//s ` `v_2 = 50m//s` `A_i = 2xx 10^(-3)m`. `A_r = ((v_2-v_1)/(v_2+v_1))A_i = 2/3 xx (10^-3)m ` ` A_t = ((2v_2)/(v_1 +v_2)) A_i = 8/3 xx (10^(-3))m ` . In second medium, speed becomes two times. Therefore, `lambda` also becomes two times. So, k remains one-half, value of `omega` will remain unchanged. Further, second medium is rarer medium `(v_2 gt v_1)`. Hence there is no change in phase angle anywhere. `:. y_r = 2/3 xx 10^(-3) cos pi (2.0x + 50t)` and `y_t = 8/3 xx 10^(-3) cos pi (2.0x + 50t)` .

Discussion

32.	A 60.0 cm guitar string under a tension of 50.0 N has a mass per unit length of 0.100 `g//cm.` What is the highest resonance frequency of the string that can be heard by a person able to hear frequencies upto 20000 Hz?
Answer» Correct Answer - A Fundamental frequency, `f= v/(2l) = sqrt(T//mu)/(2l)` `= sqrt((50)//(0.1 xx 10^-3//10^-2)/(2 xx 0.6)` ` = 58.96 Hz` Let, n th harmonic is the hightest frequency, then (58.93)n = 20000 `:. N= 339.38` Hence, 339 is the highest frequency. `:. f_(max) = (339)(58.93) Hz = 19977 Hz.`

Discussion

Explore topic-wise InterviewSolutions in .

Among two interfering sources, let `S_1` be ahead of the phase by `90^@` relative to `S_2` . If an observation point P is such that `PS_1 - PS_2 = 1.5 lambda` , the phase difference between the waves from `S_1 and S_2` reaching P isA. `3pi`B. `(5pi)/2`C. `(7pi)/2`D. `4pi`

A string that is 10 cm long is fixed at both ends. At t=0, a pulse travelling from left to right at `1cm//s is 4.0 cm ` from the right end as shown in figure. Determine the next two times when the pulse will be at that point again. State in each case whether the pulse is upright or inverted . .

A standing wave is formed by two harmonic waves, `y_1 = A sin (kx-omegat) and y_2 = A sin (kx + omegat)` travelling on a string in opposite directions. Mass density energy between two adjavent nodes on the string.

Which of the following equations can form stationary waves? (i)`y= A sin (omegat - kx)` (ii) `y= A cos (omegat - kx)` (iii) `y= A sin (omegat + kx)` (iv) `y= A cos (omegat - kx)` .A. (i) and (ii)B. (i) and (iii)C. (iii) and (iv)D. (ii) and (iv)

Regarding stationary waves, choose the correct options.A. This is an example of interferenceB. Amplitudes of waves may be differentC. particles at nodes are always at restD. Energy is conserved

The tension, length, diameter and density of a string B are double than that of another string A. Which of the following overtones of B is same as the fundamental frequency of A?A. 1stB. 2ndC. 3rdD. 4th

When tension of a string is increased by 2.5 N, the initial frequency is altered in the ratio of 3:2. The initial tension in the string isA. 6NB. 5NC. 4ND. 2N

The lengths of two wires of same material are in the ratio 1:2, their tensions are in the ratio 1:2 and their diameters are in the ratio 1:3. the ratio of the notes they emits when sounded together by the same source isA. `sqrt(2)`B. `sqrt3`C. `2sqrt(3)`D. `3sqrt(2)`

A string 1m long is drawn by a 300 Hz vibrator attached to its end. The string vibrates in three segments. The speed of transverse waves in the string is equal toA. `100 m//s`B. `200 m//s`C. `300 m//s`D. `400 m//s`

Sound waves of frequency `600 H_(Z)` fall normally on perfectly reflecting wall. The distance from the wall at which the air particles have the maximum amplitude of vibration is (speed of sound in air = `330 m//s`)

The ratio of intensities of two waves is 9:16. If these two waves interfere, then determine the ratio of the maximum and minimum possible intensities.

Two identical harmonic pulses travelling in opposite directions in a taut string approach each other. At the instant when they completely overlap, the total energy of the string will be .A. zeroB. partly kinetic and partly potentialC. purely kineticD. purely potential

The period of oscillations of a point is 0.04 s and the velocity of propagation of oscillation is 300 `m//s`. The difference of phases between the oscillations of two points at distance 10 m and 16 m respectively from the source of oscillations isA. `2pi`B. `pi/2`C. `pi/4`D. `pi`

A wire fixed at both ends is 1.0 m long and has a mass of 36g. One of its oscillation frequencies is `250 Hz` and the next higher one is 300 Hz. (a) Which harmonics do these frequencies represent? (b) What is the tension in the wire?

Two different stretched wires have same tension and mass per unit length. Fifth overtone frequency of the first wire is equal to second harmonic frequency of the second wire. Find the ratio of their lengths.

Find the fundamental frequency and the next three frequencies that could cause a standing wave pattern on a string that is 30.0 m long, has a mass per unit length of `9.00 xx 10^(-3) kg//m` and is stretched to a tension of 20.0 N.

In terms of T, `mu` and l, find frequency of (a) fourth overtone mode (b) third harmonic .

If the frequencies of the sound and fifth harmonics of a string differ by 54 Hz. What is the fundamental frequency of the string ?

For a certain stretched string, three consecutive resonance frequencies are observed as 105, 175 and 245 Hz respectively. Then, the fundamental frequency isA. 30 HzB. 45 HzC. 35 HzD. None of these

The harmonic wave `y_i = (2.0 xx 10^(-3)) cos pi (2.0x - 50t)` travels along a string towards a boundary at x=0 with a second string. The wave speed on the second string is `50 m//s`. Write expressions for reflected and transmitted waves. Assume SI units.

A 60.0 cm guitar string under a tension of 50.0 N has a mass per unit length of 0.100 `g//cm.` What is the highest resonance frequency of the string that can be heard by a person able to hear frequencies upto 20000 Hz?