Question

plz give me answer it's is very urgent
QUE NO. 3 ​

Answer 1

we know that, distance between the consecutive rarefaction/compression of a sound wave is its wavelength.

wavelength = 30 cm = 30÷100 m = 0.3 m

Hence, wavelength of the sound wave is 0.3 m

Hence, given–

$wavelength \: ( \gamma ) = 30 \: cm = 30 \times {10}^{ - 2} \: m = 0.3 \: cm \\ velocity \: (v) = 360 \: m {s}^{ - 1 }$

now, we know that

velocity = frequency x wavelength

frequency = velocity ÷ wavelength

$v = \frac{360}{0.3} = \frac{3600}{3} \\ v = 1200 \: hz$

Hence, frequency of the sound wave is 1200 Hz.

now,

frequency = 1/time period

hence,

time period = 1/frequency

$t = \frac{1}{v } \\ t = \frac{1}{1200} \: sec$

Hence, wavelength, frequency and time period of the given sound wave are 0.3 m, 1200 Hz and 1/1200 s respectively.