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 speed of the transverse wave in the string under two different conditions: when the car is at rest and when it is accelerating.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Speed of wave when the car is at rest, \( V_1 = 60 \, \text{cm/s} = 0.6 \, \text{m/s} \)
- Speed of wave when the car is accelerating, \( V_2 = 62 \, \text{cm/s} = 0.62 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Understanding Wave Speed in the String:**
The speed of a wave in a string is given by the formula:
\[ V = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the string and \( \mu \) is the linear mass density of the string.

3. **Case 1: Car at Rest**
- When the car is at rest, the tension \( T \) in the string is equal to the weight of the ball:
\[ T = mg \]
Thus, the wave speed is:
\[ V_1 = \sqrt{\frac{mg}{\mu}} \]

4. **Case 2: Car Accelerating**
- When the car is accelerating, a pseudo force acts on the ball due to the acceleration \( a \) of the car. The tension \( T' \) in the string can be expressed as:
\[ T' = \sqrt{(mg)^2 + (ma)^2} \]
Therefore, the wave speed becomes:
\[ V_2 = \sqrt{\frac{T'}{\mu}} = \sqrt{\frac{\sqrt{(mg)^2 + (ma)^2}}{\mu}} \]

5. **Setting Up the Equations:**
From the two cases, we have:
\[ V_1^2 = \frac{mg}{\mu} \]
\[ V_2^2 = \frac{\sqrt{(mg)^2 + (ma)^2}}{\mu} \]

6. **Dividing the Equations:**
Dividing the second equation by the first:
\[ \frac{V_2^2}{V_1^2} = \frac{\sqrt{(mg)^2 + (ma)^2}}{mg} \]

7. **Substituting Values:**
Substituting the known values:
\[ \frac{(0.62)^2}{(0.6)^2} = \frac{\sqrt{(mg)^2 + (ma)^2}}{mg} \]
\[ \frac{0.3844}{0.36} = \frac{\sqrt{(mg)^2 + (ma)^2}}{mg} \]
Simplifying gives:
\[ 1.0677 = \sqrt{1 + \left(\frac{a}{g}\right)^2} \]

8. **Squaring Both Sides:**
Squaring both sides:
\[ 1.139 = 1 + \left(\frac{a}{g}\right)^2 \]
Rearranging gives:
\[ \left(\frac{a}{g}\right)^2 = 0.139 \]

9. **Finding Acceleration \( a \):**
\[ a = g \sqrt{0.139} = 10 \sqrt{0.139} \]
Calculating:
\[ a \approx 10 \times 0.372 = 3.72 \, \text{m/s}^2 \]

### Final Answer:
The acceleration of the car is approximately \( 3.72 \, \text{m/s}^2 \).