Question

An unknown source plays a pitch of middle C (262 Hz). How fast would the sound wave from this source have to travel to raise
the pitch to C sharp (271 Hz)? Use 343 m/s as the speed of sound.
a. 2.8 m/s
c. 11.4 m/s
b. 15.6 m/s
d. 8.2 m/s

Answer 1

Answer:

The sound travels at $v_{s}=11.4 \mathrm{m} / \mathrm{s}$

Option: c

Explanation:

Unknown source plays of middle C (fs) = 262 Hz

The sound wave from this source have to travel to raise the pitch to C sharp is (fd) = 272 Hz

$\begin{array}{l}{velocity of sound in air(v)=343 \mathrm{m} / \mathrm{s}, f_{\mathrm{s}}=262 \mathrm{Hz}, f_{\mathrm{d}}=271 \mathrm{Hz}} \\ {velocity of receiver(v_{\mathrm{d}})=0 \mathrm{m} / \mathrm{s},velocity of source( v_{\mathrm{s}}) \text { is unknown }}\end{array}$

$\text { Speed of sound } \mathrm{V}_{\mathrm{S}}=343 \mathrm{m} / \mathrm{s}$

$f_{\mathrm{d}}=f_{\mathrm{s}}\left(\frac{v-v_{\mathrm{d}}}{v-v_{\mathrm{s}}}\right)$

$\frac{f_{d}}{f_{s}}=\left(\frac{v-v_{d}}{v-v_{s}}\right)$

$\left(v-v_{s}\right)=\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)$

$v_{s}=v-\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)$

Substitute the given values in the formula,

$v_{s}=343+\frac{262}{271}(343-0)$

$v_{s}=343+0.966(343)$

$v_{s}=343-331.33$

$v_{s}=11.4 \mathrm{m} / \mathrm{s}$

Therefore, The sound travels at $v_{s}=11.4 \mathrm{m} / \mathrm{s}$