A heavy ball is suspended from the ceiling of a motor car through a light string. A transverse pulse travels at a speed of `60 cm s^-1` on the string when the car is at rest and `62 cm s^-1` when the car accelerates on a horizontal road. Find the acceleration of the car. Take `g = 10 ms^-2`

Explanation

To solve the problem, we need to find the acceleration of the car based on the change in the speed of a transverse wave in a string when the car accelerates. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the relationship between wave speed and tension
The speed of a transverse wave \( V \) on a string is given by the formula:
\[ V = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the string and \( \mu \) is the mass per unit length of the string. Since \( \mu \) remains constant, any change in wave speed will be due to a change in tension.

### Step 2: Set up the equations for the two scenarios
1. When the car is at rest:
\[ V = 60 \, \text{cm/s} = 0.6 \, \text{m/s} \quad \text{(let's denote tension as } T\text{)} \]
\[ 0.6 = \sqrt{\frac{T}{\mu}} \quad \Rightarrow \quad T = 0.6^2 \mu = 0.36 \mu \]

2. When the car is accelerating:
\[ V' = 62 \, \text{cm/s} = 0.62 \, \text{m/s} \quad \text{(let's denote new tension as } T'\text{)} \]
\[ 0.62 = \sqrt{\frac{T'}{\mu}} \quad \Rightarrow \quad T' = 0.62^2 \mu = 0.3844 \mu \]

### Step 3: Relate the tensions
The change in tension can be analyzed using the forces acting on the heavy ball. When the car accelerates, an additional pseudo force acts on the ball due to the car's acceleration \( a \). The new tension \( T' \) can be expressed as:
\[ T' = mg + ma \]
where \( m \) is the mass of the ball and \( g \) is the acceleration due to gravity.

### Step 4: Set up the equation for tension ratio
We can express the ratio of the tensions:
\[ \frac{T'}{T} = \frac{0.3844 \mu}{0.36 \mu} = \frac{0.3844}{0.36} \]
Calculating this gives:
\[ \frac{T'}{T} = \frac{0.3844}{0.36} \approx 1.067 \]

### Step 5: Use the relationship between tension and acceleration
From the previous analysis, we have:
\[ \frac{T'}{T} = \frac{mg + ma}{mg} = 1 + \frac{a}{g} \]
Setting the two expressions for the tension ratio equal gives:
\[ 1 + \frac{a}{g} = 1.067 \]
Subtracting 1 from both sides:
\[ \frac{a}{g} = 0.067 \]
Multiplying both sides by \( g \) (where \( g = 10 \, \text{m/s}^2 \)):
\[ a = 0.067 \times 10 = 0.67 \, \text{m/s}^2 \]

### Final Answer
The acceleration of the car is:
\[ \boxed{0.67 \, \text{m/s}^2} \]