A wire of length 2L, is made by joining two wires A and B of same length but different radii r and 2

Q: Why is the tension the same in both wires?

Because the wires are connected in series and are in static equilibrium horizontally. If the tension were different, the joint would accelerate horizontally, which doesn't happen.

Q: Why do both wires vibrate with the same frequency?

They are part of a single coupled system driven by the same source. The boundary conditions force them to oscillate in sync to maintain continuity at the joint.

Q: What does it mean that the joint is a node?

A node is a point of zero amplitude. This implies that the standing waves in section A and section B perfectly match the boundary condition of zero displacement at their shared boundary, allowing us to treat them as two separate strings fixed at both ends.

Q: Why does the thicker wire have more antinodes?

The thicker wire has a higher mass density, which slows down the wave speed. Since $v = f \lambda$ and $f$ is constant, a lower speed means a shorter wavelength. A shorter wavelength fits more half-waves (antinodes) into the same length $L$.

Q: Would the ratio change if the lengths were different?

Yes! If the lengths were $L_A$ and $L_B$, the equation would be $\frac{p v_A}{2L_A} = \frac{q v_B}{2L_B}$. The ratio $p/q$ would then depend on both the ratio of the speeds and the ratio of the lengths.

Question

A wire of length $2L$, is made by joining two wires $A$ and $B$ of same length but different radii $r$ and $2r$ and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire $A$ is $p$ and that in $B$ is $q$, then the ratio $p:q$ is

Options:

(A) 3 : 5

(B) 4 : 9

(C) 1 : 2

(D) 1 : 4

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Wires

Imagine a composite wire stretched tightly between two rigid walls. This wire is made of two distinct sections, $A$ and $B$, joined perfectly in the middle. Both sections share the exact same length $L$ and are forged from the identical material. However, they differ in thickness: wire $A$ has a radius $r$, while wire $B$ is twice as thick with a radius $2r$.

When this composite string is plucked or driven by an external oscillator, it vibrates. The problem presents a fascinating constraint: the joint between the two wires acts as a **node**. This means the joint remains perfectly stationary, effectively decoupling the standing wave patterns in wire $A$ and wire $B$, even though they are vibrating at the exact same frequency $f$.

### Mass Density and Wave Speed

To understand how the waves behave in each section, we first need to determine their linear mass densities ($\mu$). The linear mass density is the mass per unit length, which can be expressed as the volume density ($\rho$) multiplied by the cross-sectional area ($A = \pi r^2$).

For wire $A$:
$$ \mu_A = \rho \pi r^2 = \mu $$

For wire $B$, the radius is doubled ($2r$), which quadruples the cross-sectional area:
$$ \mu_B = \rho \pi (2r)^2 = 4\rho \pi r^2 = 4\mu $$

Now, let's look at the wave speed. The speed of a transverse wave on a stretched string is governed by the tension $T$ and the linear mass density $\mu$:
$$ v = \sqrt{\frac{T}{\mu}} $$

Since the wires are connected in series, the tension $T$ must be uniform throughout the entire length to maintain equilibrium. Let's compare the wave speeds:
$$ v_A = \sqrt{\frac{T}{\mu}} $$
$$ v_B = \sqrt{\frac{T}{4\mu}} = \frac{1}{2} \sqrt{\frac{T}{\mu}} = \frac{v_A}{2} $$

The wave travels exactly half as fast in the thicker wire $B$!

### Standing Waves and Frequencies

Because the joint is a node, both wire $A$ and wire $B$ act like independent strings fixed at both ends. The frequency of the $n$-th harmonic for a string of length $L$ is given by:
$$ f = \frac{n v}{2L} $$

We are told that wire $A$ forms $p$ antinodes (loops) and wire $B$ forms $q$ antinodes. Since they are driven as a single coupled system, their frequencies must be identical ($f_A = f_B$).

Let's write out the frequency equations for both sections:
$$ f_A = \frac{p v_A}{2L} $$
$$ f_B = \frac{q v_B}{2L} $$

### The Final Ratio

Equating the two frequencies gives us the master equation for this problem:
$$ \frac{p v_A}{2L} = \frac{q v_B}{2L} $$

The $2L$ terms gracefully cancel out, leaving a direct relationship between the number of loops and the wave speeds:
$$ p v_A = q v_B $$

Now, we substitute our earlier finding that $v_B = \frac{v_A}{2}$:
$$ p v_A = q \left( \frac{v_A}{2} \right) $$

The $v_A$ terms cancel out, yielding:
$$ p = \frac{q}{2} \implies \frac{p}{q} = \frac{1}{2} $$

The ratio of the number of antinodes in wire $A$ to wire $B$ is **$1:2$**. This makes perfect physical sense: the thicker wire has a slower wave speed, which results in a shorter wavelength for the same frequency. A shorter wavelength naturally accommodates more loops within the same length $L$.

Select Answer:

Visualized Solution

Visualizing the Setup

Mass per Unit Length $μ$

Ratio of Linear Mass Densities

Wave Speed $v$

Ratio of Wave Speeds

Standing Wave Frequencies $f$

Equating Frequencies $f_{A} = f_{B}$

Final Calculation for $p : q$

The Way Forward

The Sigma Insight: Standing Waves in Strings and Organ Pipes

The Setup

Mass Density and Wave Speed

Standing Waves and Frequencies

The Final Ratio

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