Q.A steel wire has a length of 12.0m and a mass of 2.10kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20∘C=343m s−1?
Kerala DhseTextbookSubjective· 3mImportance★★★★★est
16% · 9/58 Questions
✓ Free question
Concept understanding — Wave Speed on String
Wave Speed on a String – From Intuition to Formula
Imagine you and a friend hold a long, taut rope between you. If you give your end a quick flick upward, a bump travels along the rope toward your friend. That bump is a wave, and the speed at which it moves is the wave speed.
Now ask yourself: what determines how fast that bump travels? Two things stand out from everyday experience:
Tension – If you pull the rope tighter, the bump zips along faster. A loose rope makes the wave crawl.
Mass – If the rope is heavy (like a thick clothesline), the wave moves slower than on a light, thin string under the same tension.
So wave speed increases with tension and decreases with the "heaviness" of the string. That's the core intuition.
The Precise Statement
For a wave traveling along a stretched string, the wave speed v is given by:
v=μT
where:
T is the tension in the string (in newtons, N)
μ is the linear mass density – the mass per unit length of the string (in kg/m)
v=μT
This formula is exact for an ideal string (perfectly flexible, no stiffness, no damping). It comes from solving the wave equation for a string, but you can understand it physically.
Why the Square Root? A Quick Physical Argument
Think of a small segment of the string. The tension provides the restoring force that tries to straighten the string when it's bent. A higher tension means a stronger restoring force, so the wave accelerates faster – hence higher speed.
The mass per unit length μ is the inertia of the string. A heavier string resists acceleration more, so the wave slows down.
The square root appears because the relationship between force, mass, and acceleration isn't linear when you derive it properly. But the key takeaway is:
Important
Wave speed on a string depends only on the string's tension and its linear density – not on the frequency or amplitude of the wave.
This is a surprising and important result. Whether you send a slow, gentle ripple or a fast, sharp pulse, both travel at the same speed on the same string.
A Simple Example
A steel guitar string has μ=0.002kg/m and is under tension T=100N. What is the wave speed?
v=0.002100=50000≈224m/s
That's about half the speed of sound in air – fast enough that the wave reaches the other end almost instantly.
Common Mistakes to Avoid
Watch out
Do not confuse wave speed with the speed of the string's particles. The string itself moves up and down (transverse motion), but the wave travels horizontally. These are different speeds.
Wave speed does NOT depend on frequency. Changing how fast you flick your hand changes the frequency, but the wave still travels at v=T/μ.
Tension is not the same as force applied at the end. If the string is under tension T everywhere (ideal case), that's the value you use – not the force you apply to create the wave.
Where This Formula Comes From (A Glimpse)
If you're curious, the derivation uses Newton's second law on a tiny curved segment of the string. For small displacements, the net vertical force from tension equals μΔx times the acceleration. This leads to the wave equation:
∂t2∂2y=μT∂x2∂2y
Comparing with the standard wave equation ∂t2∂2y=v2∂x2∂2y gives v2=T/μ, hence v=T/μ.
Note
For exams, you only need to remember and apply the formula v=T/μ. The derivation is for understanding, not memorization – unless your syllabus explicitly asks for it.
Quick Summary
Quantity
Symbol
Effect on wave speed
Tension
T
Higher tension → faster wave
Linear density
μ
Heavier string → slower wave
Frequency
f
No effect
Amplitude
A
No effect
Final takeaway: Wave speed on a string is determined entirely by the string's material and how tightly it's stretched. It's a property of the medium, not the wave itself.
Many students find this page while searching "Wave Speed on String formula physics" or "Wave Speed on String important questions and answers"; the concept sits firmly within the Class 11 Physics NCERT/CBSE syllabus. It's also a frequent building block for numericals in JEE Main, NEET and state engineering/medical entrance exams, so treating it as a one-time memorisation task rather than an understood idea tends to backfire later.
Concept: Wave speed on a string depends on tension and linear mass density: v=T/μ.
Step 1: Find the linear mass density μ of the wire.
μ=lengthmass=12.0m2.10kg=0.175kg/m.
Step 2: Set the wave speed equal to the given speed of sound: v=343m/s.
Step 3: Use v=T/μ and solve for tension T:
T=μv2=0.175×(343)2.
Step 4: Compute: 3432=117649, so T=0.175×117649=20588.575N.
✓Final answer
The required tension is 2.06×104N (or 20.6kN).
The wave speed on a string depends only on tension and linear mass density. We find the linear density from the given mass and length, then solve for the tension that makes the wave speed equal to 343 m/s. The required tension is about 2.06×104N.
The speed of a transverse wave on a stretched string is a beautiful example of how a simple mechanical property — tension — controls wave propagation. The formula is clean and intuitive: a tighter string (more tension) makes waves travel faster; a heavier string (more mass per unit length) slows them down. Here, we’re told the string’s total mass and length, so we can find its linear mass density. Then we set the wave speed equal to the given speed of sound and solve for the tension.
Let’s go step by step.
Find the linear mass densityμ of the wire.
Linear mass density is mass per unit length:
μ=lengthmass=12.0m2.10kg=0.175kg/m.
Recall the wave speed formula for a transverse wave on a string under tension T:
v=μT.
This comes from Newton’s second law applied to a small segment of the string — the restoring force is proportional to tension, and the inertia is proportional to μ.
Set the wave speed equal to the given speedv=343m/s and solve for T:
343=0.175T.
Square both sides to remove the square root:
3432=0.175T.
Multiply through by 0.175 to isolate T:
T=0.175×3432.
Calculate:
3432=117649,
T=0.175×117649=20588.575N.
Rounding to three significant figures (since the given data has three significant figures: 12.0 m, 2.10 kg, 343 m/s), we get:
T≈2.06×104N.
Watch out
A common mistake is to forget that the wave speed formula uses linear mass density, not total mass. Always divide mass by length first. Also, don’t confuse this with the speed of sound in the wire material itself — that’s a different concept (bulk modulus vs. tension).
Tip
Notice that the tension here is enormous — over 20,000 N. That’s because steel wire is heavy (high μ), so to make waves travel as fast as sound in air, you need a huge pull. In practice, such a wire would be near its breaking point.
✓Final answer
The required tension is 2.06×104N.
Step 1: Linear mass density μ=m/L=2.10/12.0=0.175 kg/m.
Step 2: Wave speed formula v=T/μ; set v equal to the target speed 343 m/s and solve for T: T=μv2.
Step 3: Substitute: T=0.175×3432=0.175×117649≈2.06×104 N.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
KEAM 2026Set eng-2026-04174 marksMCQ
Q.In the given wave equation y=0.05sinλ2π(x−200t)m, the velocity of the wave (in ms−1) is
(A) 2200
(B) 400
(C) 2002
(D) 2300
(E) 200
›Reveal solutionSolution
The wave form (x−200t) shows the wave speed directly: v=200ms−1.
The standard travelling-wave form is y=Asink(x−vt), where v is the wave speed. The given equation is:
y=0.05sinλ2π(x−200t)
Comparing the bracket (x−200t) with (x−vt) directly gives:
v=200ms−1
✓Final answer
The correct option is (E).
KEAM 2026Set eng-2026-04184 marksMCQ
Q.The speed of the transverse waves on a steel wire of length 50 cm and mass 5 g subjected to a tension of 64 N is
(A) 6.4ms−1
(B) 810ms−1
(C) 80ms−1
(D) 6.4ms−1
(E) 640ms−1
›Reveal solutionSolution
Linear mass density μ=0.01kgm−1, so wave speed v=T/μ=64/0.01=80ms−1.
Speed of a transverse wave on a stretched wire:
v=μT,μ=Lm
With m=5g=0.005kg and L=0.5m:
μ=0.50.005=0.01kgm−1
With tension T=64N:
v=0.0164=6400=80ms−1
✓Final answer
The correct option is (C).
KEAM 2026Set eng-2026-04194 marksMCQ
Q.The equation of a transverse wave in a string, y=3sin2π(25t+0.4x) m. The wavelength of the wave is
(A) 4.5 m
(B) 3 m
(C) 2.5 m
(D) 3.5 m
(E) 6.5 m
›Reveal solutionSolution
Read the coefficient of x: 2π(0.4)=2π/λ⇒λ=1/0.4=2.5 m.
The wave is y=3sin2π(25t+0.4x), so the phase is 2π(25t+0.4x).
Comparing with the standard form y=Asin2π(ft+λx), the coefficient of x gives
λ1=0.4⇒λ=0.41=2.5m.
✓Final answer
The correct option is (C).
KEAM 2025Set eng-2025-04264 marksMCQ
Q.If the speed of the transverse wave in a wire under certain tension T is v, then its speed under tension 2T (in ms−1) is
(A) 2v
(B) 2v
(C) 2v
(D) 23v
(E) 2v
›Reveal solutionSolution
The speed of a transverse wave on a wire is v=T/μ, so it scales as T. Doubling the tension multiplies the speed by 2.
Formula:
v=μT
The linear mass density μ is unchanged, so
vv′=T2T=2
v′=2v
✓Final answer
The correct option is (C).
KEAM 2024Set eng-2024-06074 marksMCQ
Q.Speed of a transverse wave on a stretched string under tension T and linear density μ is
(A) Tμ
(B) μT
(C) μT
(D) μT
(E) Tμ
›Reveal solutionSolution
Wave speed depends on tension and linear mass density as v=T/μ.
For a stretched string under tension T with linear density μ,
v=μT
✓Final answer
The correct option is (B).
KEAM 2024Set pha-2024-06104 marksMCQ
Q.The velocity of a travelling plane wave given by $y = 10^{-2} \sin \left200t - \frac{x}{5} \right m, $ is
(A) $10$ ms^{-1}
(B) $500$ ms^{-1}
(C) $400$ ms^{-1}
(D) $5$ ms^{-1}
(E) $1000$ ms^{-1}
›Reveal solutionSolution
Read ω and k from y=10−2sin(200t−x/5).
Here ω=200 rad/s and k=1/5 m−1. Wave speed:
v=kω=1/5200=1000 m/s.
✓Final answer
The correct option is (E). v=1000 m/s.
KEAM 2021Set eng-2021-P1-A14 marksMCQ
Q.The velocity of a transverse wave propagating on a stretched string represented by the equation, y=0.5sin(2πt+3πx) is (where x and y are in metres and t in seconds)
(A) 0.5ms−1
(B) 1.0ms−1
(C) 2ms−1
(D) 3ms−1
(E) 1.5ms−1
›Reveal solutionSolution
The wave speed is v=ω/k=1.5ms−1.
Concept and Intuition
For a wave y=Asin(ωt±kx), the magnitude of the propagation speed is v=ω/k, the ratio of the time and space angular frequencies.
Step-by-Step Solution
Identify ω=2πrad/s and k=3πrad/m.
v=kω=π/3π/2=23=1.5ms−1.
Common Mistakes
Taking v=ωk or reading amplitude 0.5 as the speed.