A long wire PQR is made by joining two wires PQ and QR of equal radii. PQ has length 4.8 m and mass

Q: Why does the reflected wave have a negative amplitude?

The negative sign indicates a phase inversion of $\pi$ radians. This happens whenever a wave traveling in a rarer medium (higher speed, lower density) reflects off a denser medium (lower speed, higher density).

Q: Does the transmitted wave ever undergo a phase change?

No, the transmitted wave never undergoes a phase change at a boundary, regardless of whether it is moving from a rarer to a denser medium or vice versa. Its amplitude always retains the same sign as the incident wave.

Q: Why is the tension uniform throughout both segments of the wire?

Since the composite wire is in static equilibrium and we neglect any friction or external horizontal forces, the tension must be uniform across the entire length to balance out at every point.

Q: Is energy conserved during this reflection and transmission?

Yes, energy is fully conserved. The incident power equals the sum of the reflected power and the transmitted power ($P_i = P_r + P_t$). You can verify this using the power formula $P = \frac{1}{2} \mu v \omega^2 A^2$.

Q: What would happen if the wave traveled from the heavier wire to the lighter wire instead?

If the wave traveled from $QR$ to $PQ$, then $v_1$ would be $32\text{ m/s}$ and $v_2$ would be $80\text{ m/s}$. In this case, $v_2 > v_1$, so the reflected wave would have a positive amplitude (no phase inversion).

Question

A long wire $PQR$ is made by joining two wires $PQ$ and $QR$ of equal radii. $PQ$ has length $4.8	ext{ m}$ and mass $0.06	ext{ kg}$. $QR$ has length $2.56	ext{ m}$ and mass $0.2	ext{ kg}$. The wire $PQR$ is under a tension of $80	ext{ N}$. A sinusoidal wave pulse of amplitude $3.5	ext{ cm}$ is sent along the wire $PQ$ from the end $P$. No power is dissipated during the propagation of the wave pulse. Calculate
(a) the time taken by the wave pulse to reach the other end $R$ and
(b) the amplitude of the reflected and transmitted wave pulse after the incident wave pulse crosses the joint $Q$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a composite string stretched under a uniform tension of $T = 80	ext{ N}$. This string is made of two distinct materials joined seamlessly at a junction point $Q$.

To understand how a wave behaves as it travels across this boundary, we must first determine the physical properties of each segment. The speed of a transverse wave on a stretched string is governed by its tension $T$ and its linear mass density $\mu$ (mass per unit length):

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

Let's calculate these parameters step-by-step for both segments.

### Wave Speed in Segment $PQ$

For the first segment, $PQ$, we are given:
- Length, $L_1 = 4.8	ext{ m}$
- Mass, $m_1 = 0.06	ext{ kg}$

The linear mass density $\mu_1$ is:

$$\mu_1 = \frac{m_1}{L_1} = \frac{0.06}{4.8} = 0.0125	ext{ kg/m} = \frac{1}{80}	ext{ kg/m}$$

Now, substituting the tension $T = 80	ext{ N}$ into our wave speed formula:

$$v_1 = \sqrt{\frac{80}{1/80}} = \sqrt{6400} = 80	ext{ m/s}$$

### Wave Speed in Segment $QR$

For the second segment, $QR$, we have:
- Length, $L_2 = 2.56	ext{ m}$
- Mass, $m_2 = 0.2	ext{ kg}$

Its linear mass density $\mu_2$ is:

$$\mu_2 = \frac{m_2}{L_2} = \frac{0.2}{2.56} = 0.078125	ext{ kg/m} = \frac{1}{12.8}	ext{ kg/m}$$

Substituting this into the wave speed formula:

$$v_2 = \sqrt{\frac{80}{1/12.8}} = \sqrt{1024} = 32	ext{ m/s}$$

Notice that segment $QR$ is significantly heavier than segment $PQ$, which causes the wave speed to drop from $80	ext{ m/s}$ to $32	ext{ m/s}$.

### Part (a): Total Travel Time

The total time $t$ taken by the wave pulse to travel from end $P$ to end $R$ is simply the sum of the times spent in each segment:

$$t = t_{PQ} + t_{QR} = \frac{L_1}{v_1} + \frac{L_2}{v_2}$$

Substituting our calculated speeds:

$$t = \frac{4.8}{80} + \frac{2.56}{32} = 0.06	ext{ s} + 0.08	ext{ s} = 0.14	ext{ s}$$

Thus, the wave pulse takes **$0.14	ext{ s}$** to traverse the entire composite wire.

### Part (b): Reflection and Transmission at the Boundary

When a wave pulse encounters a boundary where the wave speed changes, it cannot simply pass through unaffected. To maintain physical continuity at the junction $Q$, two conditions must be satisfied:
1. **Continuity of Displacement:** The string cannot break; therefore, the displacement just to the left of the junction must equal the displacement just to the right.
2. **Continuity of Slope:** The string cannot have a sharp kink, which would imply an infinite acceleration of an infinitesimal mass element at the boundary.

Applying these boundary conditions yields the standard equations for the reflected amplitude $A_r$ and transmitted amplitude $A_t$ in terms of the incident amplitude $A_i$:

$$A_r = \left(\frac{v_2 - v_1}{v_2 + v_1}ight) A_i$$

$$A_t = \left(\frac{2 v_2}{v_2 + v_1}ight) A_i$$

Let's calculate these amplitudes using $A_i = 3.5	ext{ cm}$, $v_1 = 80	ext{ m/s}$, and $v_2 = 32	ext{ m/s}$.

#### Reflected Amplitude

$$A_r = \left(\frac{32 - 80}{32 + 80}ight) 	imes 3.5 = \left(\frac{-48}{112}ight) 	imes 3.5 = -\frac{3}{7} 	imes 3.5 = -1.5	ext{ cm}$$

The negative sign carries deep physical meaning: it indicates that the reflected wave undergoes a **phase change of $\pi$ radians** (inversion) because it reflects off a denser medium ($v_2 < v_1$).

#### Transmitted Amplitude

$$A_t = \left(\frac{2 	imes 32}{32 + 80}ight) 	imes 3.5 = \left(\frac{64}{112}ight) 	imes 3.5 = \frac{4}{7} 	imes 3.5 = 2.0	ext{ cm}$$

The transmitted wave is always in phase with the incident wave, hence its amplitude is positive (**$2.0	ext{ cm}$**).

Visualized Solution

Visualizing the Composite Wire Setup

The Physics Tool: Wave Speed on a Stretched String

Calculating Wave Speed in Segment $P Q$

Calculating Wave Speed in Segment $QR$

Part (a): Total Time Taken to Reach End $R$

Part (b): Boundary Conditions at the Junction

Calculating Reflected Wave Amplitude $A_{r}$

Calculating Transmitted Wave Amplitude $A_{t}$

The Way Forward: Energy Conservation Check

The Sigma Insight: Reflection and Transmission of Waves

Analyzing the Setup

Wave Speed in Segment $P Q$

Wave Speed in Segment $QR$

Part (a)

Part (b)

# Reflected Amplitude

# Transmitted Amplitude

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