A heavy ball of mass M is suspended from the ceiling of a car by a light string of mass m (m

Q: Why do we neglect the mass of the string when calculating tension?

The problem states that $m \ll M$. This means the mass of the string is negligible compared to the heavy ball, so the tension is uniform throughout the string and depends only on the forces acting on the heavy ball.

Q: Why does the string tilt when the car accelerates?

In the non-inertial frame of the accelerating car, a pseudo force $Ma$ acts on the ball in the direction opposite to the car's acceleration. This force, combined with gravity, creates a new equilibrium position at an angle.

Q: How did we get the effective gravity $g'$?

The effective gravity is the vector sum of the actual acceleration due to gravity $g$ (downwards) and the pseudo acceleration $a$ (backwards). Since they are perpendicular, their resultant magnitude is $\sqrt{g^2 + a^2}$.

Q: Why did we use the binomial approximation?

The ratio of the speeds is $60.5/60 = 1 + 1/120$. Since $1/120$ is much less than 1, it implies that $a^2/g^2$ is also very small. The binomial expansion $(1+x)^n \approx 1+nx$ makes it incredibly easy to solve equations with fractional powers when $x$ is small.

Q: What if we didn't use the binomial approximation?

You would have to calculate $(1 + 1/120)^4$, which is tedious without a calculator. The binomial approximation is a standard mathematical tool in physics to simplify such expressions without losing significant accuracy.

Question

A heavy ball of mass $M$ is suspended from the ceiling of a car by a light string of mass $m (m<<M)$. When the car is at rest, the speed of transverse waves in the string is $60 	ext{ ms}^{-1}$. When the car has acceleration $a$, the wave speed increases to $60.5 	ext{ ms}^{-1}$. The value of $a$, in terms of gravitational acceleration $g$ is closest to

Options:

(A) $\frac{g}{20}$

(B) $\frac{g}{5}$

(C) $\frac{g}{30}$

(D) $\frac{g}{10}$

Sigma Percentile · Accepted Answer

The journey of mastering physics is often about connecting seemingly unrelated concepts. In this beautiful problem, we are bridging the world of **Waves on a String** with the mechanics of **Non-Inertial Frames** and **Pseudo Forces**. It’s a classic JEE setup that tests not just your memory of formulas, but your ability to visualize a physical situation and apply mathematical approximations elegantly.

Let's dive into the thought process step-by-step!

### Analyzing the Setup

Imagine you are sitting inside a stationary car. Suspended from the ceiling is a heavy ball of mass $M$, hanging perfectly vertically. The string holding it has a very small mass $m$, which means we can safely ignore the string's own weight when calculating the tension.

When the car is at rest, the only forces acting on the heavy ball are gravity pulling it down and the tension in the string pulling it up. Since the ball is in equilibrium, the tension $T$ is simply equal to the weight of the ball:
$$T = Mg$$

We are given that the speed of transverse waves on this string is $60 	ext{ ms}^{-1}$. The fundamental formula for the speed of a transverse wave on a stretched string is:
$$v = \sqrt{\frac{T}{\mu}}$$
where $\mu$ is the linear mass density of the string ($\mu = m/L$).

Substituting our tension, we get our first master equation:
$$v = \sqrt{\frac{Mg}{\mu}} = 60 	ext{ ms}^{-1}$$

### The Accelerated Frame

Now, the driver steps on the gas, and the car accelerates forward with an acceleration $a$. What happens to the hanging ball?

If you observe this from inside the car (a non-inertial frame), you must apply a **pseudo force** to apply Newton's laws. This pseudo force acts in the direction opposite to the car's acceleration. So, the ball experiences a backward force of magnitude $Ma$.

Simultaneously, gravity is still pulling it down with a force $Mg$. The string will tilt backwards until the tension $T'$ balances the resultant of these two perpendicular forces.

Using vector addition, the new effective weight (or the new tension $T'$) is the hypotenuse of the right-angled triangle formed by $Mg$ and $Ma$:
$$T' = \sqrt{(Mg)^2 + (Ma)^2} = M\sqrt{g^2 + a^2}$$

Because the tension has increased, the string is stretched tighter, and naturally, the wave speed will increase. The problem states the new speed is $60.5 	ext{ ms}^{-1}$. Let's write our second equation:
$$v' = \sqrt{\frac{T'}{\mu}} = \sqrt{\frac{M\sqrt{g^2 + a^2}}{\mu}} = 60.5 	ext{ ms}^{-1}$$

### The Ratio and Binomial Magic

We have two equations and we need to find $a$. The most elegant way to eliminate the unknown $\mu$ and $M$ is to take the ratio of the two speeds:
$$\frac{v'}{v} = \frac{60.5}{60}$$

Substituting the expressions for $v'$ and $v$:
$$\frac{\sqrt{M\sqrt{g^2 + a^2}/\mu}}{\sqrt{Mg/\mu}} = \frac{60.5}{60}$$

Notice how beautifully $M$ and $\mu$ cancel out! We are left with:
$$\sqrt{\frac{\sqrt{g^2 + a^2}}{g}} = 1 + \frac{0.5}{60}$$

Let's simplify the right side: $0.5/60 = 1/120$. 
On the left side, bringing $g$ inside the inner square root makes it $g^2$:
$$\left( \frac{g^2 + a^2}{g^2} ight)^{1/4} = 1 + \frac{1}{120}$$
$$\left( 1 + \frac{a^2}{g^2} ight)^{1/4} = 1 + \frac{1}{120}$$

Now, we face a fractional power of $1/4$. Calculating this directly would be a nightmare. But physics is the art of approximation! Notice that $1/120$ is a very small number. This implies that the term $\frac{a^2}{g^2}$ must also be very small compared to 1.

Whenever we have an expression of the form $(1 + x)^n$ where $x \ll 1$, we can use the **Binomial Approximation**:
$$(1 + x)^n \approx 1 + nx$$

Applying this to our left-hand side:
$$1 + \frac{1}{4}\left(\frac{a^2}{g^2}ight) \approx 1 + \frac{1}{120}$$

### Final Calculation

The math has now collapsed into a beautifully simple linear equation. Subtract 1 from both sides:
$$\frac{1}{4}\frac{a^2}{g^2} = \frac{1}{120}$$

Multiply both sides by 4:
$$\frac{a^2}{g^2} = \frac{4}{120} = \frac{1}{30}$$

Taking the square root of both sides gives us the acceleration $a$:
$$a = \frac{g}{\sqrt{30}}$$

To find the closest option, we need to estimate $\sqrt{30}$. We know that $5^2 = 25$ and $6^2 = 36$. Since 30 is almost exactly in the middle, $\sqrt{30}$ is approximately $5.47$.

Therefore:
$$a \approx \frac{g}{5.47}$$

Looking at our options, $\frac{g}{5}$ is the closest match.

And there you have it! By combining wave mechanics, pseudo forces, and a clever mathematical approximation, we've cracked a seemingly complex problem. Always remember, when changes are small, the binomial theorem is your best friend!

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Visualized Solution

$v = \frac{T}{μ}$

$Accelerated Frame$

$v^{'} = \frac{T ^{'}}{μ}$

$\frac{v ^{'}}{v}$

$Simplification$

$Binomial Expansion$

$Solving for a$

$Final Approximation$

The Sigma Insight: Wave Equation and Wave Speed

Analyzing the Setup

The Accelerated Frame

The Ratio and Binomial Magic

Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v=μT​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Accelerated Frame

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v′=μT′​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vv′​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplification

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Binomial Expansion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for a

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Approximation