LinkedIn Insight Solutions for Exercises of Superposition of Waves - Grad Plus

Solutions for Exercises of Superposition of Waves

1. Choose the correct option.

i) When an air column in a pipe closed at one end vibrates such that three nodes are formed in it, the frequency of its vibrations is ….. times the fundamental frequency.

(A) 2

(B) 3

(C) 4

(D)5

ii) If two open organ pipes of length 50 cm and 51 cm sounded together produce 7 beats per second, the speed of sound is.

(A) 307 m/s

(B) 327m/s

(C) 350m/s

(D) 357m/s

iii) The tension in a piano wire is increased by 25%. Its frequency becomes ….. times the original frequency.

(A) 0.8

(B) 1.12

(C) 1.25

(D) 1.56

iv) Which of the following equations represents a wave travelling along the y-axis?

(A) x = A sin (ky — ωt)

(B) y = A sin (kx — ωt)

(C) y = A sin (ky) cos(ωt)

(D) y = A cos (ky) sin(ωt)

v) A standing wave is produced on a string fixed at one end with the other end free. The length of the string

(A) must be an odd integral multiple of λ/4.

(B) must be an odd integral multiple of λ/2.

(C) must be an odd integral multiple of λ.

(D) must be an even integral multiple of λ.


2. Answer in Brief.

i) A wave is represented by an equation y = A sin (Bx + CO . Given that the constants A, B and C are positive, can you tell in which direction the wave is moving?

ii) A string is fixed at the two ends and is vibrating in its fundamental mode. It is known that the two ends will be at rest. Apart from these is there any position on the string which can be touched so as not to disturb the motion of the string? What will be the answer to this question if the string is vibrating in its first and second overtones?

iii) What are harmonics and overtones?

iv) For a stationary wave set up in a string having both ends fixed, what is the ratio of the fundamental frequency to the second harmonic?

v) The amplitude of a wave is represented by $latexy=0.2\sin4\pi\lbrack\frac t{0.08}-\frac x{0.8}\rbrack$ in SI units. Find (a) wavelength, (b) frequency and (c) amplitude of the wave.


3. State the characteristics of progressive waves.


4. State the characteristics of stationary waves.


5. Derive an expression for equation of stationary wave on a stretched string.


6. Find the amplitude of the resultant wave produced due to interference of two waves given as yi= Al sin wt y2 = A2 sin (ωt + ∅)


7. State the laws of vibrating strings and explain how they can be verified using a sonometer.


8. Show that only odd harmonics are present in the vibrations of air column in a pipe closed at one end.


9. Prove that all harmonics are present in the vibrations of the air column in a pipe open at both ends.


10. A wave of frequency 500 Hz is travelling with a speed of 350 m/s.
(a) What is the phase difference between two displacements at a certain point at times 1.0 ms apart? (b) what will be the smallest distance between two points which are 45° out of phase at an instant of time?

n = 500Hz, v = 350m/s a) Φ = ? for points 1 ms apart

We have,$latex \lambda\;=\;\frac cn$

$latex i.e.\;\lambda=\;\frac{350}{500}\;=\;0.7m$

As phase difference of 2π corresponds to distance λ, let phase difference Φ corresponds to distance x, then

$latex \varphi=\frac{2\pi}\lambda\times\;x$

Given, t = 1 ms =10-3s

∴ x = v × t = 350 × 10-3m

$latex \therefore\;\varphi=\frac{2\pi}{0.7}\times350\times10^{-3}=\pi$

b) x = ? for θ = 45As,

$latex \varphi=\frac{2\pi}\lambda\times\;x$

we have,

$latex \therefore\;x=\frac\lambda{2\pi}\times\;\phi$

$latex x\;=\;\frac{0.7}{2\pi}\;\times\;45\;\times\frac\pi{180}$

$latex x\;=\;0.0875m\;=\;8.75m$


11. A sound wave in a certain fluid medium is reflected at an obstacle to form a standing wave. The distance between two successive nodes is 3.75 cm. If the velocity of sound is 1500 m/s, find the frequency.

???? = 3.75cm = 3.75 × 10-2m, v=1500 m/s,

n=?

We have for fundamental node,

$latex n\;=\;\frac v{2l}$

$latex n\;=\;\frac{1500}{2\;\times\;3.75\;\times\;10^{-2}}$

$latex n\;=\;20000\;Hz\;=\;20\;KH$


12. Two sources of sound are separated by a distance 4 m. They both emit sound with the same amplitude and frequency (330 Hz), but they are 180° out of phase. At what points between the two sources, will the sound intensity be maximum?
[Ans: ± 0.25, ± 0.75, ± 1.25 and ±1.75 m from the point at the center]


13. Two sound waves travel at a speed of 330 m/s. If their frequencies are also identical and are equal to 540 Hz, what will be the phase difference between the waves at points 3.5 m from one source and 3 m from the other if the sources are in phase?

v = 330m/s, n = 540Hz, x=3.5m, x2=3m

ΔΦ=?

We know that,

$latex \phi=\frac{2\pi}\lambda\times\;x=\frac{2\pi}{\displaystyle v/n}\times\;x=\frac{2\pi n}v\times\;x$

$latex \therefore\;\triangle\phi\;=\;\phi_1\;-\;\phi_2$

$latex \therefore\;\triangle\phi\;=\frac{2\pi n}v(x_1-x_2)$

$latex \therefore\;\triangle\phi\;=\;\phi_1\;-\;\phi_2$

$latex \therefore\;\triangle\phi\;=\;\frac{2\pi n}v(x_1\;-\;x_2)$

$latex \therefore\;\triangle\phi\;=\;\frac{1080}{330}\;\times\;0.5$

$latex \therefore\triangle\phi=\frac{1080}{330}\times0.5$

$latex \therefore\triangle\phi=1.64\pi$


14. Two wires of the same material and same cross section are stretched on a One wire is loaded with 1.5 kg and another is loaded with 6 kg. The vibrating length of first wire is 60 cm and its fundamental frequency of vibration is the same as that of the second wire. Calculate vibrating length of the other wire.

T = 1.5kg, T = 6kg, l = 60cm = 60 × 10-2m,

n1 = n2, m1 = m2, l2=?

The fundamental frequency for the vibration is, given as

$latex n\;=\;\frac1{2l}\sqrt{\frac Tm}n$

As, n1 = n2 we have

$latex \frac1{2l_1}\sqrt{\frac{T_1}{m_1}}=\frac1{2l_2}\sqrt{\frac{T_2}{m_2}}$

$latex \therefore\frac{l_2}{l_1}=\sqrt{\frac{T_2}{T_1}}$

$latex \therefore l_2=l_1\sqrt{\frac{T_2}{T_1}}$

$latex \therefore l_2=60\times10^{-2}\times\sqrt{\frac6{1.5}}$

$latex \therefore l_2=60\times10^{-2}\times\;2$

$latex \therefore l_2=120\times10^{-2}\;m=1.2m$


15. A pipe closed at one end can produce overtones at frequencies 640 Hz, 896 Hz and 1152 Hz. Calculate the fundamental frequency.

nx = 640Hz, nx+1 = 896Hz and nx+2 = 1152Hz n = ?

We know that for pipe closed at one end, frequency of the overtone is odd multiple of fundamental frequency, hence we have,

n1 = 3n = (2(1) + 1)n,

n2 = 5n = (2(2) + 1)n,

n3 = 7n = (2(3) + 1)n and likewise.

Hence we can observe that,

n2 − n1 = n3 − n2 = 2n

That means difference in the frequencies of any two consecutive overtones is 2n i.e. two times the fundamental frequency. In our question, we have, difference in the frequencies of any two consecutive overtones

= (896 − 640) = (1152 − 896) = 256

∴ 2n = 256 ∴ n = 128Hz


16. A standing wave is produced in a tube open at both ends. The fundamental frequency is 300 Hz. What is the length of tube? (speed of the sound = 340 m s-1).

v = 340m/s, n = 300 Hz, l = ?

We have for fundamental node,

$latex n=\frac v{2l}$

$latex l\;=\;\frac v{2n}\;=\;\frac{340}{2\times300}$

$latex \therefore\;l\;=\;0.57m$


17. Find the fundamental, first overtone and second overtone frequencies of a pipe, open at both the ends, of length 25 cm if the speed of sound in air is 330 m/s.

???? = 25cm = 25 × 10-2m, v = 330m/s,

a) Fundamental frequency for the pipe open at both ends,

$latex n=\frac v{2l}$

$latex\therefore\;n\;=\;\frac{330}{2\times25\times10^{-2}}$

$latex \therefore\;n\;=\;660Hz$

b) 1st overtone frequency for the pipe open at both ends,

n1 = 2 × n = 2 × 660 = 1320Hz

c) 2nd overtone frequency for the pipe open at both ends,

n2 = 3 × n = 3 × 660 = 1980Hz


18. A pipe open at both the ends has a fundamental frequency of 600 Hz. The first overtone of a pipe closed at one end has the same frequency as the first overtone of the open pipe. How long are the two pipes?

For pipe open at both ends, n = 600Hz

Given that, (n1)closed = (n1)open For the pipe open at both ends, fundamental frequency is given as, 600Hz. But,

$latex n_{open}=\frac v{2l}$

$latex\therefore\;l\;=\;\frac v{2n}$

$latex \therefore\;l\;=\frac{330}{2\times600}$

$latex \therefore\;l\;=\;0.275m\;=\;27.5cm$

For the pipe closed at one end, 1st overtone frequency is same as 1st overtone frequency of open pipe.

∴ (n1)closed = (n1)open = 2n = 1200 But,

$latex {(n_1)}_{closed}=\frac{3v}{4l}$

$latex \therefore\frac{3v}{4l}=1200$

$latex \therefore\;l\;=\;\frac{3\;\times\;330}{4\;\times\;1200}$

$latex \therefore\;l\;=\;0.20625m$

∴ l = 20.625cm


19. A string lm long is fixed at one end. The other end is moved up and down with frequency 15 Hz. Due to this, a stationary wave with four complete loops, gets produced on the string. Find the speed of the progressive wave which produces the stationary wave.[Hint: Remember that the moving end is an antinode.]

l = 1m, n = 15 Hz,

Given that 4 complete loops are produced but as there is antinode at the moving end total 9/2 loops are produced.

Now we also know that one loop corresponds to λ/2.

Hence, 9/2 loops corresponds to

$latex \frac92\;\times\;\frac\lambda2\;=\;\frac{9\lambda}4$

$latex \therefore\frac{9\lambda}4=1$

$latex \therefore\lambda=\frac{4\times1}9=\frac49$

$latexv=\;n\lambda\;=\;15\;\times\;\frac49$

$latex \therefore\;v\;=\;6.67m/s$


20. A violin string vibrates with fundamental frequency of 440Hz. What are the frequencies of first and second overtones?

A violin string is an example of a pipe open at both ends.

n1 = Frequency of 1st ovetone = 2n = 2 × 440

∴ n1 = 880Hz

& n2 = Frequency of 2nd ovetone = 3n = 3 × 440

∴ n2 = 1320Hz


21. A set of 8 tuning forks is arranged in a series of increasing order of frequencies. Each fork gives 4 beats per second with the next one and the frequency of last fork is twice that of the first. Calculate the frequencies of the first and the last

λ1 = 400 nm = 400× 10-9 m,

λ2 = 700 nm = 700 × 10-9 m, μg = 1.55

λ1 = ?and λ2 = ? in glass

We know,

$latex \mu_g\;=\;\frac{\lambda_a}{\lambda_g}$

$latex \therefore\;\lambda_g\;=\;\frac{\lambda_a}{\mu_g}$

$latex \therefore\;\lambda_{1g}\;=\;\frac{\lambda_1}{\mu_g}$

$latex \therefore\;\lambda_{1g}\;=\;\frac{400\;\times\;10^{-9}}{1.55}$

$latex =258.06\times10^{-9}$

$latex \lambda_{2g}\;=\;\frac{\lambda_2}{\mu_g}$

$latex \lambda_{2g}\;=\;\frac{900\;\times\;10^{-9}}{1.55}$

$latex \lambda_{2g}\;=\;451.61\;\times\;10^{-9}$

Hence range is 258.06 nm to 451.61 nm.


22. A sonometer wire is stretched by tension of 40 N. It vibrates in unison with a tuning fork of frequency 384 Hz. How many numbers of beats get produced in two seconds if the tension in the wire is decreased by 1.24 N?

T1 = 40N, n1 = 384 Hz,

Number of beats = ? in t = 2s, T2 = 40 – 1.24 = 38.76N

For beat frequency we will need (n2 − n1)

For Sonometer experiment we have,

$latex \frac{n_1}{\sqrt{T_1}}=\frac{n_2}{\sqrt{T_2}}$

$latex \frac{384}{\sqrt{40}}\;=\;\frac{n_2}{\sqrt{38.76}}$

$latex \therefore\;n_2\;=\;384\sqrt{\frac{38.76}{40}}$

$latex \therefore n_2=378\;Hz\;(u\sin g\;\log)$

∴ Beat frequency = n1 – n2

∴ Beat frequency = 384 – 378

∴ Beat frequency = 6 Hz

∴ Number of beats heard in 2s = 6 × 2 = 12 beats


23. A sonometer wire of length 0.5 m is stretched by a weight of 5 kg. The fundamental frequency of vibration is 100 Hz. Calculate linear density of wire.

l = 0.5m, w = 5 kg, n = 100 Hz, m = ?

We know that,

$latex n=\frac1{2l}\sqrt{\frac Tm}=\frac1{2l}\sqrt{\frac{wg}m}$

$latex \therefore\;100\;=\;\frac1{2\times0.5}\sqrt{\frac{5\times9.8}m}$

$latex \therefore\;100\;=\;\sqrt{\frac{49}m}$

$latex\therefore\;10000\;=\frac{49}m$

$latex \therefore\;m\;=\;\frac{49}{10^4}$

$latex \therefore\;m\;=\;49\;\times\;10^{-4}$

$latex \therefore\;m\;=\;4.9\;\times\;10^{-3}kg/m$


24. The string of a guitar is 80 cm long and has a fundamental frequency of 112 Hz. If a guitarist wishes to produce a frequency of 160 Hz, where should the person press the string?

l = 80cm = 80 × 10-2m, n1 = 112 Hz,

l =? if n2 = 160 Hz

We know that,

According to law of length, n1l1 = n2l2

$latex \therefore\;I_2\;=\;\frac{n_1l_1}{n_2}$

$latex \therefore\;I_2\;=\;\frac{112\;\times\;80\;\times\;10^{-2}}{160}$

$latex \therefore\;I_2\;=56\times10^{-2}$

$latex \therefore\;l_2\;=\;56cm\;$


 

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