Q.A steel wire 0.72 m long has a mass of 5.0×10−3 kg. If the wire is under a tension of 60 N, what is the speed of transverse waves on the wire?
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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:
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=0.002 kg/m and is under tension T=100 N. What is the wave speed?
v=0.002100=50000≈224 m/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
- 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) …
Concept: v=T/μ, μ=m/L.
μ=0.725.0×10−3≈6.944×10−3 kg/m. v=6.944×10−360=8640.6≈93 m/s. …
Using v=T/μ with the wire's linear mass density μ=m/L, the speed of transverse waves on the steel wire is v≈93 m/s.
The governing formula
The speed of a transverse wave on a stretched wire depends on the tension T and its linear mass density μ:
v=μT
Step-by-step calculation
1. Find the linear mass density.
μ=Lm=0.725.0×10−3≈6.944×10−3 kg/m
2. Identify the tension.
T=60 N.
3. Apply the wave speed formula.
v=μT=6.944×10−360=8640.6≈92.95 m/s≈93 m/s …
Step 1: Linear mass density μ=Lm=0.725.0×10−3≈6.944×10−3 kg/m.
Step 2: Transverse wave speed on a stretched wire: v=T/μ. …
- KEAM 2026Set eng-2026-04174 marksMCQQ.In the given wave equation y=0.05 sinλ2π(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=200 ms−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) …
- KEAM 2026Set eng-2026-04184 marksMCQQ.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: …
- KEAM 2026Set eng-2026-04194 marksMCQQ.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 …
- KEAM 2025Set eng-2025-04264 marksMCQQ.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 …
- KEAM 2024Set eng-2024-06074 marksMCQQ.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, …
- KEAM 2024Set pha-2024-06104 marksMCQQ.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: …
- KEAM 2021Set eng-2021-P1-A14 marksMCQQ.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=1.5 ms−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=2π rad/s and k=3π rad/m. …
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