A man claps his hands in front of a wall and he hears an echo after 1.6s. He walks. 33m towards wall and hears the echo 1.4 s after clapping. Then velocity of sound in air is:
330 m/s
Calculating the Velocity of Sound Using Echo Times
This problem involves using the concept of an echo to determine the velocity of sound in air. An echo occurs when sound waves reflect off a surface and return to the source. The time taken to hear the echo is the time for the sound to travel from the source to the reflector and back again.
Understanding the Echo Principle
When a person claps in front of a wall, the sound travels to the wall and reflects back. The total distance covered by the sound in hearing the echo is twice the distance between the person and the wall. If the distance to the wall is $d$ and the time taken to hear the echo is $t$, the velocity of sound $v$ can be calculated using the formula:
\( v = \frac{\text{Total Distance}}{\text{Time}} \)
Since the total distance is \(2d\), the formula becomes:
\( v = \frac{2d}{t} \)
Setting Up the Equations
We are given two scenarios:
- Initial scenario: Man is at a distance \(d_1\) from the wall, and the echo time is \(t_1 = 1.6\) s.
- Second scenario: Man walks \(33\) m towards the wall, so the new distance is \(d_2 = d_1 - 33\). The new echo time is \(t_2 = 1.4\) s.
Let \(v\) be the velocity of sound in air (which remains constant).
Using the formula \( v = \frac{2d}{t} \), we can write two equations based on the two scenarios:
Equation 1 (Initial):
\( v = \frac{2d_1}{t_1} \)
\( v = \frac{2d_1}{1.6} \)
\( 2d_1 = 1.6v \) --- (A)
Equation 2 (After moving):
\( v = \frac{2d_2}{t_2} \)
\( v = \frac{2(d_1 - 33)}{1.4} \)
\( 2(d_1 - 33) = 1.4v \)
\( 2d_1 - 66 = 1.4v \) --- (B)
Solving for Velocity of Sound
Now we have a system of two linear equations with two variables, \(d_1\) and \(v\).
- \( 2d_1 = 1.6v \)
- \( 2d_1 - 66 = 1.4v \)
Substitute the expression for \(2d_1\) from Equation (A) into Equation (B):
\( (1.6v) - 66 = 1.4v \)
Now, we solve for \(v\):
\( 1.6v - 1.4v = 66 \)
\( 0.2v = 66 \)
\( v = \frac{66}{0.2} \)
\( v = \frac{660}{2} \)
\( v = 330 \text{ m/s} \)
The calculated velocity of sound in air is \(330\) m/s.
Comparing with Options
Let's compare our result with the given options:
- 320 m/s
- 350 m/s
- 330 m/s
- 340 m/s
Our calculated value of \(330\) m/s matches Option 3.
| Parameter | Initial Scenario | Second Scenario |
|---|---|---|
| Distance to Wall | \(d_1\) | \(d_2 = d_1 - 33\) m |
| Echo Time | \(t_1 = 1.6\) s | \(t_2 = 1.4\) s |
| Total Distance by Sound | \(2d_1\) | \(2d_2 = 2(d_1 - 33)\) |
| Velocity of Sound | \(v\) | \(v\) |
The problem demonstrates how changes in distance and time related to an echo can be used to determine the speed of sound, assuming constant velocity.
Revision Table: Echo Calculation
| Concept | Formula Used | Application in Problem |
|---|---|---|
| Echo Distance | Distance traveled by sound = \(2 \times\) Distance to reflector | \(2d_1\) in 1.6s, \(2(d_1-33)\) in 1.4s |
| Velocity, Distance, Time | \(v = \frac{\text{Distance}}{\text{Time}}\) | \(v = \frac{2d_1}{1.6}\), \(v = \frac{2(d_1-33)}{1.4}\) |
| Solving Simultaneous Equations | Using substitution or elimination | Solving for \(v\) using the two velocity equations |
Additional Information: Velocity of Sound in Air
The velocity of sound in air is not constant under all conditions. It is primarily affected by temperature. At standard atmospheric pressure, the speed of sound in dry air can be approximated by the formula:
\( v \approx (331.3 + 0.606 \times T) \text{ m/s} \)
where \(T\) is the temperature in degrees Celsius.
- Humidity also has a small effect, increasing the speed slightly.
- Air pressure has very little effect on the speed of sound in air at typical atmospheric pressures.
- The frequency or wavelength of sound does not affect its speed in a given medium.
The value calculated in this problem, 330 m/s, is close to the speed of sound in dry air at 0°C (approx 331.3 m/s) or slightly below, depending on the exact temperature.
Important Questions from The Speed of Sound
An object is 10.64 km below the sea level. A research team sends down a sonar signal to confirm this depth. After how long can it expect to get the echo? Take speed of sound in sea water = 1520 m/s.
A boy stands 83 m in front of a high wall and then blows a whistle, calculate the time interval when he hears an echo. Speed of sound is 332 m/s.
For hearing a distinct sound, the time interval between the original sound and the reflected one must be atleast ______ seconds.
The speed of a longitudinal wave in a solid bar is given by v = √(X/ρ), where 'ρ' is density of the medium. What is the unknown term 'X'?
In which medium the speed of sound will be maximum?
