The extension in a string, obeying Hooke's law, is x. The speed of transverse wave in the stretched

Q: Why doesn't the mass per unit length change when the string is stretched?

In standard introductory physics problems, we assume the extension is small enough that the change in the string's linear mass density is negligible. However, in reality, stretching a string does decrease its linear mass density slightly.

Q: Can I use $v = f \lambda$ to solve this?

While $v = f \lambda$ is always true, it doesn't help here because we don't know how the frequency or wavelength changes without first knowing how the physical properties of the medium (tension and density) change.

Q: What is Hooke's law constant $k$ in this context?

Hooke's law constant $k$ represents the stiffness of the string. It depends on the material's Young's modulus and cross-sectional area, and it remains constant during the stretching process.

Q: Why is the wave speed proportional to the square root of tension rather than tension itself?

This comes from the derivation of the wave equation using Newton's second law on a small string segment. The restoring force restoring the string to equilibrium scales with tension, while the inertia scales with mass, leading to the square root relationship.

Q: What would happen if the extension was doubled instead of increased to 1.5x?

If the extension was doubled ($2x$), the tension would double ($2T$). The new wave speed would then be $\sqrt{2}v \approx 1.41v$.

Question

The extension in a string, obeying Hooke's law, is $x$. The speed of transverse wave in the stretched string is $v$. If the extension in the string is increased to $1.5x$, the speed of transverse wave will be

Options:

(A) $1.22 v$

(B) $0.61 v$

(C) $1.50 v$

(D) $0.75 v$

Sigma Percentile · Accepted Answer

### Introduction to Wave Propagation on Stretched Strings

Imagine holding a tightly stretched guitar string.

When you pluck it, a wave travels down its length, bouncing back and forth to create beautiful music.

But what dictates how fast that wave travels?

In this article, we will dive deep into the physics of wave speed on elastic strings, exploring the beautiful interplay between **Hooke's Law** and **wave mechanics**.

---

### Analyzing the Setup

Let's look at the problem at hand.

We are given a string that obeys **Hooke's Law**.

Hooke's Law states that the tension $T$ in an elastic material is directly proportional to its extension $x$:

$$T = kx$$

Here, $k$ is the force constant of the string.

This means that if we stretch the string further, the tension inside it increases linearly.

Initially, the string has an extension of $x$, which produces a certain tension $T_1 = kx$.

Under this tension, a transverse wave travels along the string with a speed $v$.

---

### The Master Equation for Wave Speed

The speed of a transverse wave on a stretched string is determined by two physical properties: the tension $T$ pulling it tight, and the mass per unit length $\mu$ (linear mass density) resisting motion due to inertia.

The relationship is given by the famous formula:

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

Since the mass of the string is constant and we assume the change in length doesn't significantly alter the linear mass density $\mu$ for small extensions, we can establish a direct proportionality:

$$v \propto \sqrt{T}$$

Now, combining this with Hooke's Law ($T \propto x$), we get a beautiful and simple relationship between the wave speed and the extension of the string:

$$v \propto \sqrt{x}$$

This is our master key to solving the problem!

---

### Calculating the New Wave Speed

Let's set up a ratio between the final state (State 2) and the initial state (State 1).

In the initial state:
- Extension $x_1 = x$
- Wave speed $v_1 = v$

In the final state:
- Extension $x_2 = 1.5x$
- Wave speed $v_2 = v'$

Using our proportionality relation:

$$\frac{v_2}{v_1} = \sqrt{\frac{x_2}{x_1}}$$

Substituting the values into this ratio:

$$\frac{v'}{v} = \sqrt{\frac{1.5x}{x}}$$

The extension variable $x$ cancels out perfectly:

$$\frac{v'}{v} = \sqrt{1.5}$$

Now, we evaluate the square root of $1.5$:

$$\sqrt{1.5} \approx 1.2247$$

Therefore, the new wave speed is:

$$v' \approx 1.22 v$$

This matches **Option (a)** perfectly!

---

### Real-World Considerations: The Way Forward

In textbook physics, we often assume that the linear mass density $\mu$ remains constant during stretching.

But what happens in reality?

As you stretch a real elastic string, its length increases while its total mass remains constant.

This means the mass per unit length $\mu = \frac{M}{L}$ actually **decreases** as the extension increases!

Since $v = \sqrt{\frac{T}{\mu}}$, a decrease in $\mu$ would make the wave speed increase even faster than our calculated $1.22v$.

This is a classic example of how real-world engineering and physics go beyond simplified models to reveal even richer behavior!

Animated Solution for Physics - Waves: The extension in a string, obeying Hooke's law, is $x$ . The speed of transverse wave in the stretched string is $v$ . If the extension in the string is increased to $1.5 x$ , the speed of transverse wave will be

Select Answer:

Visualized Solution

Visualizing the Stretched String

Hooke's Law and Tension

Speed of Transverse Waves

Relating Speed to Extension

Setting up the Ratio

Substituting the Values

Calculating the Final Speed

Exploring Further Variations

The Sigma Insight: Wave Equation and Wave Speed

Introduction to Wave Propagation on Stretched Strings

Analyzing the Setup

The Master Equation for Wave Speed

Calculating the New Wave Speed

Real-World Considerations

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