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?

Concept & Intuition

In a wire under tension, two types of mechanical waves travel at different speeds:

• Transverse waves depend on tension (stress × area) and linear density.
• Longitudinal waves depend on the material’s elasticity (Young’s modulus) and density.

The problem gives a ratio between these speeds and Young’s modulus, so we can relate stress to known quantities. The trick is to express both speeds in terms of stress and density, then cancel density using the ratio.

Step-by-step solution

1. Write the speed formulas For a wire under tension $T$ with cross-sectional area $A$ and linear mass density $\mu$:
   • Speed of transverse wave:

• Speed of longitudinal wave:

 where $Y$ is Young’s modulus and $\rho$ is volume density.

2. Relate linear density to volume density

Since $\mu = \rho A$, we can rewrite the transverse speed as:

where $\sigma = T/A$ is the stress in the wire.

3. Use the given ratio The problem states:

Substitute the expressions:

4. Cancel density Both sides have $\sqrt{1/\rho}$, so they cancel:

Square both sides:

5. Plug in Young’s modulus Given $Y = 10^{10} \, \text{Pa}$:

Concept & Intuition

A sonometer string obeys the law: frequency $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$, where $L$ is length, $T$ tension, $\mu$ linear mass density (constant here). So $f \propto \frac{\sqrt{T}}{L}$. If you halve $L$, frequency would double if tension stayed same. But here frequency halves, so tension must drop significantly — specifically, by a factor that compensates both the length change and the frequency change.

Step-by-step reasoning

1. Write the relation for original case Original frequency: $f_1 = 400\ \text{Hz}$, length $L_1 = L$, tension $T_1 = T$.

2. Write the relation for new case New frequency: $f_2 = 200\ \text{Hz}$, length $L_2 = \frac{L}{2}$, tension $T_2 = T'$.

3. Take the ratio of the two equations

4. Plug in the known frequencies

So:

5. Solve for the tension ratio Square both sides:

Hence new tension : original tension = $1 : 16$.

Concept & Intuition

When you send identical wave pulses from one end of a clamped string, each pulse travels to the other end, reflects (inverted because the end is fixed), and comes back. If you send a second pulse just as the first returns, the two pulses will meet and overlap. For constructive interference, the pulses must arrive in phase — meaning the second pulse should be launched exactly when the first returns, so their displacements add. The minimum time between pulses that achieves this is simply the round-trip travel time.

1. Find the wave speed on the string The speed of a transverse wave on a string under tension is given by

where $T = 2.5\ \mathrm{N}$ is the tension and $\mu$ is the mass per unit length.

The string has mass $m = 10^{-3}\ \mathrm{kg}$ and length $L = 0.25\ \mathrm{m}$, so

Thus

2. Determine the round-trip time A pulse travels from one clamped end to the other (distance $L = 0.25\ \mathrm{m}$) in time

To go to the far end and return to the starting point takes twice that:

3. Why this is the minimum for constructive interference If you send a pulse at $t=0$, it returns at $t=0.02\ \mathrm{s}$. Sending the next pulse exactly at that moment means the two pulses meet at the starting end, both moving in the same direction (the returning one is now moving toward the other end again). They are in phase because the reflected pulse has undergone a phase inversion at the fixed end, but the second pulse hasn’t reflected yet — however, the condition for constructive interference here is that the displacements add when they overlap. The simplest way to guarantee this is to have the second pulse start exactly when the first returns, so they travel together. Any shorter interval would cause partial cancellation.

For the fundamental of a string, $f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$, so $f\propto\dfrac{1}{L}\sqrt{T}$. Thus

With $L_2=2L_1$, $f_1=200$, $f_2=300$:

So new tension : original tension $=9:1$.

Concept: a string fixed at both ends vibrating in its n-th normal mode (n-th harmonic) has

L = n (lambda/2) and f_n = n v / (2L)

Fourth normal mode: n = 4, L = 2 m, f = 500 Hz.

lambda = 2L/n = 2(2)/4 = 1 m

v = f lambda = 500 x 1 = 500 m/s

(Equivalently v = 2Lf/n = 2 x 2 x 500 / 4 = 500 m/s.)

ANSWER: C

Concept: a harmonic wave y = A cos(omega t + kz) has amplitude A (the coefficient of the cosine).

Given y = 20 cos(omega t + 4z): amplitude A = 20 (and wave number k = 4).

ANSWER: B

Standard form: $y=A\sin2\pi\!\left(\dfrac{t}{T}-\dfrac{x}{\lambda}\right)$.

Given $y=5\sin2\pi\!\left(\dfrac{t}{0.05}-0.05x\right)$:

• $T=0.05\,$s $\Rightarrow$ frequency $f=\dfrac1T=20\,$Hz.
• $\dfrac{1}{\lambda}=0.05 \Rightarrow \lambda=20\,$m.