Question

Two identical loudspeakers 2.0 m apart are emitting sound waves into a room where the speed of sound is 340 m/sec. John is standing 5.0m in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. What is the lowest possible frequency of sound for which this is possible?

Answer 1

Answer: The lowest possible frequency of sound for which this is possible is 212.5 Hz.

Step-by-step explanation:

It is known that formula for path difference is as follows.

`\Delta L = (n + (1)/(2)) * (\lambda)/(2)` ... (1)

where, n = 0, 1, 2, and so on

As John is standing perpendicular to the line joining the speakers. So, the value of
`L_(1)` is calculated as follows.

`L_(1) = \sqrt{(2)^(2) + (5)^(2)}\\= 5.4 m`

Hence, path difference is as follows.

`\Delta L = (5.4 - 5) m = 0.4 m`

For lowest frequency, the value of n = 0.

`\Delta L = (0 + (1)/(2)) * (\lambda)/(2) = (\lambda)/(4)`

`\lambda = 4 \Delta L`

where,

`\lambda` = wavelength

The relation between wavelength, speed and frequency is as follows.

`\lambda = (\\u)/(f)\\4 \Delta L = (\\u)/(f)\\`

where,

`\\u` = speed

f = frequency

Substitute the values into above formula as follows.

`f = (\\u)/(4 \Delta L)\\f = (340)/(4 * 0.4 m)\\= 212.5 Hz`

Thus, we can conclude that the lowest possible frequency of sound for which this is possible is 212.5 Hz.