Q.The description of a wave traveling along a string is described by the equation y(x, t) = 0.005 sin(80.0x - 3.0t), where the numerical constants are in S.I. units: 0.005 m, 80.0 rad/m and 3.0 rad/s. Calculate
(a) amplitude
(b) wavelength
(c) time period of the wave.
OR
Define the following -
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.
Comparing the given wave equation with the standard form y(x,t) = a sin(kx − ωt) directly gives the amplitude, wave number, and angular frequency, from which the wavelength and time period follow. …
For y(x,t) = 0.005 sin(80x − 3t): amplitude = 0.005 m, wavelength ≈ 0.0785 m, time period ≈ 2.09 s.
This item includes an OR alternative; the primary part is answered below (a full worked numerical problem).
The standard equation of a progressive (travelling) wave moving along +x is y(x, t) = a sin(kx − ωt), where a is the amplitude, k is the angular wave number (k = 2π/λ), and ω is the angular frequency (ω = 2π/T = 2πf). Comparing this with the given equation, y(x, t) = 0.005 sin(80.0x − 3.0t):
(a) Amplitude: matching the coefficient in front of the sine function directly,
Q.Speed of transverse vibration in stretched string is
(A) √(T/m)
(B) T/m
(C) Tm
(D) m/T
›Reveal solutionSolution
The speed of a transverse wave on a stretched string is v=T/m.
For a string under tension T with linear mass density (mass per unit length) m, applying Newton's second law to a small string element in a transverse disturbance gives the wave equation with wave speed:
Q.A 10 m long steel wire has mass 5 g. If the wire is under a tension of 80 N, the speed of transverse waves on the wire is
(a) 100 ms^-1
(b) 200 ms^-1
(c) 400 ms^-1
(d) 500 ms^-1
›Reveal solutionSolution
Wave speed on a stretched string is v = sqrt(T/mu); computing the linear mass density from the given length and mass, then substituting, gives v = 400 m/s.
Length of wire, L = 10 m. Mass, m = 5 g = 5 x 10^-3 kg.
Linear mass density, mu = m/L = (5 x 10^-3 kg) / (10 m) = 5 x 10^-4 kg/m.
Q.With increase in tension in a string, frequency of transverse vibration
(A) increases
(B) decreases
(C) remains constant
(D) first increases then decreases
›Reveal solutionSolution
Frequency of a vibrating string increases with tension.
For a string of length L and mass per unit length μ under tension T, the fundamental frequency of transverse vibration is f=2L1μT. Since f∝T, increasing the tension incre …
Q.Write down the formula for the speed of transverse waves in a stretched string.
›Reveal solutionSolution
The speed of a transverse wave on a string is v=T/μ — it increases with tension and decreases with a heavier string.
Step 1 — Physical basis.
A transverse wave on a stretched string is a restoring-force phenomenon: tension T provides the restoring force that pulls a displaced element of the string back, while the string's inertia (its mass per unit length μ) resists that acceleration. A wave equation derived from Newton's second law applied to a small string element under tension gives:
v=μT
Step 2 — Meaning of the symbols.
T = tension in the string (in newtons)
μ=m/L = mass per unit length of the string (in kg/m)
…
Q.Write the relation between angular frequency (w), angular wave number (k) and wave velocity (v).
›Reveal solutionSolution
The wave velocity equals the angular frequency divided by the angular wave number: v = ω/k.
For a progressive wave y = A sin(kx - ωt), a point of constant phase (kx - ωt = constant) moves with the wave. Differentiating, k(dx/dt) - ω = 0, so dx/dt = ω/k. This dx/dt is exactly the wave (phase) velocity v. Also, since ω = 2πf (f = frequency) and k = 2π/λ (λ = wavelength), ω/k = (2πf)/(2π/λ) = fλ, which is the familiar relation v = fλ. So both forms — v = ω/ …