A pendulum made of a uniform wire of cross-sectional area A has time period T. When an additional ma

Q: Why don't we consider the initial mass of the bob when calculating $\Delta L$?

The initial length $L$ already includes the stretch due to the initial mass. $\Delta L$ is the *additional* stretch caused solely by the *additional* mass $M$.

Q: Does the mass of the bob affect the time period of a simple pendulum?

In an ideal simple pendulum with an inextensible string, mass does not affect the time period. However, here the wire is elastic, so adding mass increases the length, which in turn increases the time period.

Q: Why did we square the time periods?

Squaring the time periods removes the square root, making it much easier to algebraically manipulate the lengths and isolate the Young's modulus term.

Q: What is the restoring force in this pendulum?

The restoring force for the pendulum's swing is still the tangential component of gravity. The elastic force of the wire simply determines the equilibrium length of the pendulum.

Q: Can we assume $\Delta L$ is very small compared to $L$?

Yes, for typical metal wires, the elastic stretch is small. However, the problem doesn't require an approximation; the algebraic derivation holds exactly for the new equilibrium length.

Question

A pendulum made of a uniform wire of cross-sectional area $A$ has time period $T$. When an additional mass $M$ is added to its bob, the time period changes $T_M$. If the Young's modulus of the material of the wire is $Y$, then $\frac{1}{Y}$ is equal to
($g = $ gravitational acceleration)

Options:

(A) $\left[ \left( \frac{T_M}{T} \right)^2 - 1 \right] \frac{A}{Mg}$

(B) $\left[ \left( \frac{T_M}{T} \right)^2 - 1 \right] \frac{Mg}{A}$

(C) $\left[ 1 - \left( \frac{T_M}{T} \right)^2 \right] \frac{A}{Mg}$

(D) $\left[ 1 - \left( \frac{T}{T_M} \right)^2 \right] \frac{A}{Mg}$

Sigma Percentile · Accepted Answer

### The Setup: A Stretchy Pendulum

Imagine a simple pendulum swinging back and forth. We usually assume the string or wire holding the bob is perfectly rigid and inextensible. But in reality, materials have elasticity!

In this problem, we have a pendulum with a uniform wire of cross-sectional area $A$. Initially, it has a time period $T$. We know the formula for the time period of a simple pendulum is:

$$T = 2\pi \sqrt{\frac{L}{g}}$$

Now, we add an extra mass $M$ to the bob. Because the wire is elastic, this extra weight pulls down on the wire, causing it to stretch by a small amount, let's call it $\Delta L$. The new, longer length of the pendulum is $L + \Delta L$. This increase in length naturally leads to a new, longer time period, $T_M$:

$$T_M = 2\pi \sqrt{\frac{L + \Delta L}{g}}$$

### Enter Young's Modulus

To figure out exactly how much the wire stretches, we need to use a property of the material called **Young's Modulus ($Y$)**. Young's Modulus is a measure of stiffness and is defined as the ratio of longitudinal stress to longitudinal strain.

$$Y = \frac{	ext{Stress}}{	ext{Strain}}$$

The additional stress on the wire is caused by the added weight $Mg$ acting over the cross-sectional area $A$:

$$	ext{Stress} = \frac{Mg}{A}$$

The strain is the fractional change in length:

$$	ext{Strain} = \frac{\Delta L}{L}$$

Substituting these into the Young's Modulus equation gives us:

$$Y = \frac{\frac{Mg}{A}}{\frac{\Delta L}{L}} = \frac{MgL}{A \Delta L}$$

Rearranging this to solve for the extension $\Delta L$, we get:

$$\Delta L = \frac{MgL}{AY}$$

### The New Time Period

Now that we have an expression for the extension, let's plug it back into our equation for the new time period $T_M$:

$$T_M = 2\pi \sqrt{\frac{L + \frac{MgL}{AY}}{g}}$$

This looks a bit messy. To clean it up and relate it back to our original time period $T$, let's square both time period equations and divide the new one by the old one. This is a classic physics trick to eliminate constants like $2\pi$ and $g$!

$$\left( \frac{T_M}{T} ight)^2 = \frac{\frac{L + \Delta L}{g}}{\frac{L}{g}} = \frac{L + \Delta L}{L}$$

Simplifying the right side, we get:

$$\left( \frac{T_M}{T} ight)^2 = 1 + \frac{\Delta L}{L}$$

### Bringing It All Together

Now, let's substitute our expression for $\Delta L$ back into this ratio:

$$\left( \frac{T_M}{T} ight)^2 = 1 + \frac{\frac{MgL}{AY}}{L}$$

Notice how beautifully the original length $L$ cancels out!

$$\left( \frac{T_M}{T} ight)^2 = 1 + \frac{Mg}{AY}$$

We are almost there. The question asks us to find an expression for $\frac{1}{Y}$. Let's isolate the term containing $Y$ by subtracting 1 from both sides:

$$\frac{Mg}{AY} = \left( \frac{T_M}{T} ight)^2 - 1$$

Finally, multiply both sides by $\frac{A}{Mg}$ to leave $\frac{1}{Y}$ by itself:

$$\frac{1}{Y} = \frac{A}{Mg} \left[ \left( \frac{T_M}{T} ight)^2 - 1 ight]$$

And there we have it! By combining the kinematics of simple harmonic motion with the mechanics of solid materials, we've derived the exact relationship. This matches option (a).

Select Answer:

Visualized Solution

Initial and Final Time Periods

Young's Modulus Formula

Extension $Δ L$

Substituting $Δ L$ in $T_{M}$

Ratio of Time Periods

Simplifying the Ratio

Final Expression for $\frac{1}{Y}$

The Sigma Insight: Simple Harmonic Motion (SHM)

The Setup

Enter Young's Modulus

The New Time Period

Bringing It All Together

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