Q.A string of mass $2.50\ \text{kg}$ is under a tension of $200\ \text{N}$. The length of the stretched string is $20.0\ \text{m}$. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?

The basic wave equation states that wave speed equals the product of its wavelength and its frequency: v = lambda f.

For any periodic wave, in the time of one full period T, the wave advances forward by exactly one wavelength lambda. So:

Speed v = distance / time = lambda / T

Since frequency f = 1/T, this becomes:

v = lambda x f

This is the fundamental relation connecting a wave's speed, wavelength, and frequency -- not amplitude or period alone, which describe the wave's size and the interval of one oscillation respectively but do not by themselves determine how fast the wave pattern propagates.

The wave relation is v = f lambda, which rearranges to f = v/lambda - option (b).

The fundamental relation between the speed v, frequency f and wavelength lambda of a wave is

v = f lambda,

since a wave advances one wavelength in each time period T = 1/f. Rearranging gives

f = v/lambda.

Checking the options, only (b) f = v/lambda is consistent with v = f lambda; the others do not satisfy the relation dimensionally or algebraically.

The speed of a transverse wave on a stretched string is $v=\sqrt{T/m}$.

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:

$v = \sqrt{\dfrac{T}{m}}$

This shows the speed increases with greater tension and decreases with a heavier (denser) string.

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.

Tension, T = 80 N.

Speed of transverse waves on a stretched wire/string:

v = sqrt(T / mu) = sqrt(80 / (5 x 10^-4)) = sqrt(160000) = 400 m/s.

Frequency of a vibrating string increases with tension.

For a string of length $L$ and mass per unit length $\mu$ under tension $T$, the fundamental frequency of transverse vibration is $f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$. Since $f\propto\sqrt T$, increasing the tension increases the frequency — this is exactly why tightening a guitar or violin string raises its pitch.

The speed of a transverse wave on a stretched string is v = sqrt(T/m) (T = tension, m = linear mass density), matching option (b).

For a string under tension T with mass per unit length m (also written as mu), a transverse disturbance travels along it with speed:

v = sqrt(T/m)

This shows that a more tightly stretched string (higher T) carries waves faster, while a heavier string (higher m per unit length) carries waves more slowly -- exactly why tightening a guitar string raises the pitch, and thicker strings sound lower. Among the given options, only (b), sqrt(T/m), matches this form.

The speed of a transverse wave on a string is $v=\sqrt{T/\mu}$ — 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 $\mu$) resists that acceleration. A wave equation derived from Newton's second law applied to a small string element under tension gives:

$v = \sqrt{\frac{T}{\mu}}$

Step 2 — Meaning of the symbols.

• $T$ = tension in the string (in newtons)
• $\mu = m/L$ = mass per unit length of the string (in kg/m)

Step 3 — Interpretation.

A tighter string (larger $T$) carries waves faster; a heavier string (larger $\mu$) carries waves more slowly — exactly matching everyday observation on musical instrument strings.

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/k and v = fλ — express the same fact: the wave travels a distance of one wavelength in one time period.