A string can be used to understand the properties of a wave. Imagine, one end of the string is fixed and you are holding the other end of the string. Now, if the string is moved up and down, a wave is formed on it. The type of wave formed is called the transverse wave. A transverse wave is defined as a wave where the movement of the particles of the medium is perpendicular to the direction of the propagation of the wave. In this example, the string acts as the medium.
Download Waves on String Previous Year Solved Questions PDF
Transverse waves form peaks and troughs. The top point of the wave is called crest or peak. The bottom point of the wave is called the trough.
| Velocity of the wave on a string
Velocity is the distance travelled by the time taken. In waves, this is found by dividing the wavelength by the period: v=λ/T, where 1/T = frequency(f) |
Video Lessons:
Velocity of a Wave on a String
Standing Wave on a String
JEE Main Previous Year Solved Questions on Waves on String
Q1: A wave y = a sin(ωt – kx) on a string meets with another wave producing a node at x = 0. Then the equation of the unknown wave is
(a) y = asin(ωt + kx)
(b) y = –asin(ωt + kx)
(c) y = asin(ωt – kx)
Solution
Consider option (a)
Stationary wave: Y = asin(ωt + kx) + asin(ωt – kx)
when, x = 0, Y is not zero.
The option is not acceptable.
Consider option (b)
Stationary wave: Y = asin(ωt – kx) – asin(ωt + kx) At x = 0, Y = a sinωt – asinωt = zero
This option holds good.
Option (c) gives Y = asin(ωt – kx)+ asin(ωt – kx)
Y = 2asinωt (At x = 0, Y is not zero)
Hence only option (b) holds good
Answer: (b) y = –asin(ωt + kx)
Q2: The displacement y of a wave travelling in the x-direction is given by
y = 10-4sin(600t – 2x+ π/3) metre,
where x is expressed in metre and t in second. The speed of the wave-motion, in m s–1 is
(a) 300
(b) 600
(c) 1200
(d) 200
Solution
Given wave equation: y= 10-4 sin (600t – 2x + π/3)m
Standard wave equation: y = asin(ωt – kx + Φ)
Compare them
Angular speed, ω = 600 sec-1
Propagation constant, k = 2m-1
ω/k = (2πf)/(2π/λ) = fλ = velocity
Since velocity = ω/k = 600/2 = 300 m/sec
Answer: (a) 300
Q3: A string is stretched between fixed points separated by 75 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is
(a) 10.5 Hz
(b) 105 Hz
(c) 1.05 Hz
(d) 1050 Hz.
Solution
For the string fixed at both the ends, resonant frequency are given by f = nv/2L, where symbols have their usual meanings. lt is given that 315 Hz and 420 Hz are two consecutive resonant frequencies, let these be nth and (n + 1)the harmonics.
315 = nv/2L———(1)
420 = (n+1)v/2L —(2)
Dividing equation (1) by equation (2) we get
n = 3
From equation (1), lowest resonant frequency
f0 = v/2L = 315/3 = 105 Hz
Answer: (b) 105 Hz
Q4: The transverse displacement y(x,t) of a wave on a string is given by
This represents a
(a) wave moving in +x-direction with speed √(a/b)
(b) wave moving in –x-direction with speed √(b/a)
(c) standing wave of frequency √b
(d) standing wave of frequency 1/√b
Solution
Given the equation
Comparing equation (1) with standard equation y(x, t) = f (ax + bt)
As there is a positive sign between x and t terms, hence wave travel in –x direction.
Wave speed= Coefficient of t/Coefficient of x = √(b/a)
Answer: (b) wave moving in –x-direction with speed √(b/a)
Q5: The equation of a wave on a string of linear mass density 0.04 kg m–1 is given by
The tension in the string is
(a) 6.25 N
(b) 4.0 N
(c) 12.5 N
(d) 0.5 N
Solution
Here, linear mass density μ = 0.04 kg m–1
The given equation of a wave is
Compare it with the standard wave equation y = Asin(ωt – kx)
we get, ω=2π/0.04 rad s-1, k =2π/0.5 rad m-1
Wave velocity, v = ω/k= (2π/ 0.04)/(2π/ 0.5) ms-1 …..(1)
Also,
Here, T is the tension in the string and μ is the linear mass density
Equating equations (1) and (2), we get
T = μω2/k2
T = [ 0.04 x (2π/0.04)2] /(2π/0.5)2= 6.25 N
Answer: (a) 6.25 N
Q6: A wave travelling along the x-axis is described by the equation y(x, t) = 0.005 cos(αx –β t). If the wavelength and the time period of the wave are 0.08 m and 2.0 s, respectively, then α and β in appropriate units are
(a) α = 12.50π, β = π/2.0
(b) α = 25.00π, β = π
(c) α = 0.08/π, β = 2.0/π
(d) α = 0.04/π, β =1.0/π
Solution
The wave travelling along the x-axis is given by
y(x,t) = 0.005 cos(αx – βt)
Therefore,α = k = 2π/λ As λ = 0.08 m
α = 2π/0.08 = π/0.04 ⇒ α = (π/4 ) x 100 = 25.00π
ω = β ⇒2π/2 =π
∴ α = 25.00π
β = π
Answer: (b) α = 25.00π, β = π
Q7: A uniform string of length 20 m is suspended from a rigid support. A short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is (take g = 10 m s–2)
(a) 5π √5s
(b) 2s
(c) 2√2s
(d) √5
Solution
A uniform string of length 20 m is suspended from a rigid support.
As μ is mass per unit length of the rope, then μ =m/L
As
Integrating, on both sides, we get
Answer: (c) 2√2 s
Q8: A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1%. What is the fundamental frequency of steel if the density and elasticity of steel are 7.7 × 103 kg/m3 and 2.2 × 1011 N/m2 respectively?
(a) 770 Hz
(b) 188.5 Hz
(c) 178.2 Hz
(d) 200.5
Solution
Frequency ,f = v/2l
Where, T is tension in the wire and μ is the mass per unit length of wire.
Also, Young’s modulus, Y = Tl/AΔl
T/A = YΔl/l
Substituting this value in frequency expression, we get
Given, l = 1.5 m, Δl/l = 0.01
d= 7.7 x 103 kg/m3
Y= 2.2 x 1011 N/m2
Substituting these values we have
f = 178.2 Hz
Answer: (c) 178.2 Hz
Q9: Two sitar strings, A and B, playing the note ‘Dha’ are slightly out of tune and produce beats of frequency 5 Hz. The tension of the string B is slightly increased and the beat frequency is found to decrease by 3 Hz. If the frequency of A is 425 Hz, the original frequency of B is
(a) 428 Hz
(b) 430 Hz
(c) 420 Hz
(d) 422 Hz
Solution
Frequency of sitar string B is either 420 Hz or 430 Hz. As tension in string B is increased, its frequency will increase. If the frequency is 430 Hz, the beat frequency will increase. If the frequency is 420 Hz, the beat frequency will decrease, hence the correct answer is 420 Hz.
Answer: (c) 420 Hz
Q10: A standing wave is formed by the superposition of two waves travelling in opposite directions. The transverse displacement is given by
y(x,t)= 0.5sin[(5π/4)x]cos(200πt)
What is the speed of the travelling wave moving in the positive x-direction?
(x and t are in meter and second, respectively)
(a) 180 m/s
(b) 160 m/s
(c) 120 m/s
(d) 90 m/s
Solution
Given y(x,t)= 0.5sin[(5π/4)x]cos(200πt)
Comparing this equation with standard equation of standing wave, y(x, t) = 2a sin kx cosωt, we get, k = 5π/4 rad/m, ω = 200 rad/s
Speed of the travelling wave, v = ω/k = 200π/(5π/4) = 160 m/s
Answer: (b) 160 m/s
Q11: Length of a string tied to two rigid supports is 40 cm. Maximum length (wavelength in cm) of a stationary wave produced on it is
(a) 20
(b) 80
(c) 40
(d) 120
Solution
λmax/2 = 40 ⇒ λmax = 80 cm
Answer: (b) 80
Q12: Statement-1:Two longitudinal waves given by equations y1(x,t)=2asin(ωt – kx) and y2(x,t)=2asin(2ωt – 2kx) will have equal intensity.
Statement-2: Intensity of waves of given frequency in the same medium is proportional to square of amplitude only.
(a) Statement-1 is false, statement-2 is true
(b) Statement-1 is true, statement-2 is false
(c) Statement-1 is true, statement-2 true; statement-2 is the correct explanation of statement-1
(d) Statement-1 is true, statement-2 is true; statement -2 is not correct explanation of statement-1
Solution
y1(x,t) = 2asin(ωt – kx)
y2(x,t) = 2asin(2ωt – 2kx)
But Intensity, I = ½(ρω2A2v)
Here, ρ = density of the medium,
A= amplitude
v = velocity of the wave
Intensity depends upon amplitude, frequency and velocity of the wave.
Also, I1 = I2
Answer: (b) Statement-1 is true, statement-2 is false
Q13: A pipe open at both ends has a fundamental frequency ‘f’ in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now
(a) f
(b) f/2
(c) 3f/4
(d) 2f
Solution
When the pipe is open at both ends
λ/2 = l
λ = 2l
v = fλ
f = v/λ = v/2l ———-(1)
When the pipe is dipped vertically in the water so that half of the pipe is in water
λ/4 = l/2
λ = 2l ⇒v = f’λ
f’ = v/λ = v/2l =f——-(2)
Thus, the fundamental frequency of the air column is now,
f = f’
Answer: (a) f
Q14: A string of length 1 m and mass 5g is fixed at both ends. The tension in the string is 8.0 N. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to
(a) 10 cm
(b) 33.3 cm
(c) 16.6 cm
(d) 20.0 cm
Solution
Velocity of the wave on the string
v = 40 m/s
Here, T = tension and μ = mass/length
Wavelength of the wave λ = v/n = 40/100
Separation between successive nodes,
λ/2 = 40/(2x 100) = 20 cm
Answer: (d) 20.0 cm
Q15: Equation of a travelling wave on a stretched string of linear density 5g/m is y=0.03sin(450t – 9x) where distance and time are measured in SI units. The tension in the string is
(a) 10 N
(b) 7.5 N
(c) 12.5 N
(d) 5 N
Solution
We have given,
y = 0.03sin(450t – 9x)
Comparing it with standard equation of wave, we get
ω = 450 k=9
v = ω/k = 450/9 = 50m/s
Velocity of the travelling wave on a string is given by
μ = linear mass density
T = 2500 x 5 x 10-3
T = 12.5 N
Answer:(c) 12.5 N
Also Read:
Wave On String JEE Advanced Previous Year Questions With Solutions
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