A string of length 0.4 m and mass 10^-2 kg is tightly clamped at its ends. The tension in the string

Q: Why does the wave invert when reflecting from a clamped end?

A clamped end is a fixed boundary where displacement must always be zero. When the pulse hits this boundary, the support exerts an equal and opposite reaction force on the string, sending back an inverted pulse (phase shift of $\pi$).

Q: What would happen if one end of the string was free to move?

At a free boundary, the string can move freely, so the pulse reflects without inversion (phase shift of $0$). In that case, a single reflection would keep the pulse upright, but it would still need to travel back to the starting point to interfere.

Q: Why is the distance taken as $2L$ instead of $L$?

The pulses are generated at one end. For a returning pulse to interfere with a newly generated pulse at the same end, it must travel to the other end (distance $L$) and return (distance $L$), making the total round-trip distance $2L$.

Q: Does the shape of the pulse change during propagation?

In an ideal, non-dispersive medium, the shape of the pulse remains perfectly intact as it travels, only undergoing inversion upon reflection at fixed boundaries.

Q: Can constructive interference happen at larger time intervals?

Yes, constructive interference can also occur at integer multiples of the fundamental round-trip time, i.e., $n \times (2L/v)$. However, the question asks for the *minimum* value of $\Delta t$, which corresponds to $n=1$.

Question

A string of length $0.4	ext{ m}$ and mass $10^{-2}	ext{ kg}$ is tightly clamped at its ends. The tension in the string is $1.6	ext{ N}$. Identical wave pulses are produced at one end at equal intervals of time $\Delta t$. The minimum value of $\Delta t$, which allows constructive interference between successive pulses, is

Options:

(A) $0.05	ext{ s}$

(B) $0.10	ext{ s}$

(C) $0.20	ext{ s}$

(D) $0.40	ext{ s}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a tightly stretched string of length $L = 0.4	ext{ m}$ clamped firmly at both ends.

When we pluck or shake one end of this string, we generate a localized disturbance—a **wave pulse**—that travels down the string.

Our goal is to find the minimum time interval $\Delta t$ between successive pulses such that they interfere **constructively**.

For constructive interference to occur, the returning pulse must overlap with the newly generated pulse in the exact same phase (i.e., both must be upright or both must be inverted at the starting point).

Let's break this down into simple, logical steps.

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### Step 1: Finding the Wave Speed

The speed of a transverse wave on a stretched string depends on two physical properties: the tension $T$ pulling the string tight, and its linear mass density $\mu$ (mass per unit length).

$$\mu = \frac{M}{L}$$

Given:
- Mass of the string, $M = 10^{-2}	ext{ kg}$
- Length of the string, $L = 0.4	ext{ m}$

Substituting these values:

$$\mu = \frac{10^{-2}	ext{ kg}}{0.4	ext{ m}} = 2.5 	imes 10^{-2}	ext{ kg/m}$$

Now, we use the wave speed formula:

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

Substituting $T = 1.6	ext{ N}$ and $\mu = 2.5 	imes 10^{-2}	ext{ kg/m}$:

$$v = \sqrt{\frac{1.6}{2.5 	imes 10^{-2}}} = \sqrt{64} = 8	ext{ m/s}$$

So, any pulse generated on this string travels at a constant speed of $8	ext{ m/s}$.

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### Step 2: The Physics of Boundary Reflections

What happens when the pulse reaches the clamped end?

Because the end is tightly clamped (a **fixed boundary**), the string cannot move at that point.

To keep the displacement zero, the boundary exerts an equal and opposite reaction force on the string, causing the reflected pulse to flip upside down.

This corresponds to a **phase change of $\pi$ radians** (or $180^\circ$):

1. **First Reflection (at the far end):** The upright pulse becomes inverted (phase shift of $\pi$).
2. **Second Reflection (back at the starting end):** The inverted pulse becomes upright again (another phase shift of $\pi$).

The total phase shift after one complete round trip is:

$$\Delta \phi = \pi + \pi = 2\pi	ext{ radians}$$

A phase shift of $2\pi$ means the pulse is back in its original upright shape, perfectly ready to reinforce a newly created upright pulse!

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### Step 3: Calculating the Minimum Time Interval

Since the pulse must travel to the far end and back to return to the starting point in the correct phase, the total distance traveled is:

$$d = 2L = 2 	imes 0.4	ext{ m} = 0.8	ext{ m}$$

The time taken for this round trip is the minimum interval $\Delta t$ required for constructive interference:

$$\Delta t_{	ext{min}} = \frac{2L}{v} = \frac{0.8	ext{ m}}{8	ext{ m/s}} = 0.10	ext{ s}$$

Thus, the minimum time interval between successive pulses is **$0.10	ext{ s}$**, which corresponds to **Option (b)**.

Select Answer:

Visualized Solution

Visualizing the Setup

The Wave Speed Formula

Calculating Linear Mass Density $μ$

Calculating Wave Speed $v$

Understanding Reflection at Fixed Ends

Condition for Constructive Interference

Calculating the Minimum Time Interval $Δ t$

Final Computation

The Way Forward

The Sigma Insight: Reflection and Transmission of Waves

Analyzing the Setup

Step 1

Step 2

Step 3

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