Question

A tuning fork of frequency 220 Hz produces sound waves of wavelength 1.5m in air at S.T.P. Calculate the increase in the wavelength, when the temperature of air is 27°c.​

Answer 1

velocity of waves=

$wavelength \times frequency \\ = 1.5 \times 220 = 330$

also,

$velocity = \sqrt{ \gamma rt \div m} \\ where \: r = gas \: constant \\ \: \: \: \: \: \: \: \: \: \: \: \: \: t = temperature \\ \: \: \: \: \: \: \: \: \: \: \: \: m = mass \\ for \: a \: particuar \: gas \\ mass \: and \: \gamma = constant \\ therefore \\ velocity(0) \div velocity(27) = \\ \sqrt{temp.(0) \div temp.(27)} (in \: k) \\ = \sqrt{273 \div 300} \\ velocity(27) = 330 \times \sqrt{300 \div 273} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 345.9339 \frac{m}{s}$

therefore,

$wavelength = velocity \div frequency \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 345.9339 \div 220 = 1.572m$

Answer 2

The increased wavelength is 1.57 m .

Given:

Frequency of tuning fork = ν = 220 Hz

Wavelength of the waves from tuning fork = λ = 1.5 m

Increased temperature of tuning fork = T = 27°C

To find:

Increased wavelength = λn = ?

Formula:

Velocity of the sound wave is:

V = fλ

Solution:

Now, the velocity of the sound wave is:

V = fλ

V = 220 × 1.5

V = 330 m/s

Now, the velocity after the temperature change is:

Vn = V + 0.6T

Vn = 330 + (0.6 × 27)

Vn = 346.2 m/s

Now, the wavelength after the temperature change is:

Frequency is a property of the source (tuning fork) and the properties of the tuning fork remains the same. The frequency does not change.

Vn = f λn

346.2 = 220 λn

λn = $\frac{346.2}{220}$

λn = 1.57 m