91影视

1. The sound in air has velocity v = 343 m/s
2. Width w = 0.75 m

There would be a difference of for the distance to travel by the subsequent sound wave. Using this extra distance, find the frequency of the pulses. Also, using the same concept, compare the frequency of the pulses if the width of the terrace was smaller.

The frequency of the pulse,

$\mathbf{f}\mathbf{=}\frac{\mathbf{1}}{\mathbf{螖迟}}$

The velocity of the pulse,

$\mathbf{v}\mathbf{=}\frac{\mathbf{x}}{\mathbf{螖迟}}$

Where, v is velocity,tis time, xis displacement and f is frequency.

String of pulse returning to the stage. Extra distance covered by the pulse as compared to its previous pulse is x = 2W .so,

$鈭�t=\frac{2w}{v}$

And frequency is,

$$\begin{gathered} f=1/鈭�t \\ =\frac{v}{2\text{w}} \\ =\frac{343m/s}{2脳0.75\text{m}} \\ ==2.3脳1{\text{0}}^{\text{2}}\text{Hz} \end{gathered}$$

Hence, frequency at which the pulses return is$f=2.3脳1{\text{0}}^{\text{2}}\text{Hz}$ .

Fromtheabove equation,thefrequency is inversely proportional to the width,

$f鈭�\frac{1}{\text{w}}$

i.e. ifthewidth is smaller then frequency is higher.

Hence, if the width of the terraces were smaller, the change in frequency would be higher.