(i) A wire of length 80 cm has a frequency of 256 Hz. Calculate the length of a similar wire under similar tension, which will have frequency 1024 Hz.

(i) Given,

l₁ = 80 cm, f₁ = 256 Hz, f₂ = 1024 Hz, l₂ = ?

Since, f ∝ $\frac{1}{l}$

Therefore, f₁l₁ = f₂l₂

or

$$\begin{aligned}f_2 = \frac{f_1l_1}{f_2} \\[0.5em] = \frac{256 \times 80}{1024} \\[0.5em] = 20 \text{ cm}\end{aligned}$$

(ii) Given, f = 256 Hz, λ = 1.3 m

(a) V = fλ = 256 x 1.3 = 332.8 m s^-1

(b) Speed of second sound (V') = 332.8 m s^-1

Wavelength of second sound (λ') = 2.6 m

$$\begin{aligned}\text{f} = \frac{V'}{λ'} \\[0.5em] = \frac{332.8}{2.6} \\[0.5em] = 128 \text{ Hz}\end{aligned}$$

As frequency of second sound is less as compared to the first sound, hence, to the listener, the first sound of wavelength 1.3 m will appear to be shriller than the second sound of wavelength 2.6 m.