text: # Word Problem regarding The Doppler Effect?
  
  Jun 29, 2016
  
  $320 H z$
  
  #### Explanation:
  
  Step 1
  The problem describes a sound that is moving toward a stationary observer with an air temperature of ${15}^{\circ} C$. Since the speed of sound increases as the temperature increases, we must determine the speed of the waves at ${15}^{\circ} C$. This can be found with the formula:
  
  $\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{v}_{s} = 331 \frac{m}{s} + \left(\frac{0.6 \frac{m}{s}}{{\textcolor{w h i t e}{i}}^{\circ} C}\right) \times \text{temperature}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
  
  Using the formula, the speed of the sound waves is:
  
  $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{{v}_{s}} = 331 \frac{m}{s} + \left(\frac{0.6 \frac{m}{s}}{{\textcolor{w h i t e}{i}}^{\circ} C}\right) \times {15}^{\circ} C$
  
  $= \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{340 \frac{m}{s}}$
  
  Step 2
  Since the Doppler effect results in a change in the perceived frequency between the source and the observer, we can use the Dopper Effect formula for sound. The sign in the denominator will be a subtraction sign since the source is moving toward the observer:
  
  $\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{f}_{d} = \left({v}_{s} / \left({v}_{s} - {v}_{o}\right)\right) f} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
  
  $\underline{\text{where:}}$
  ${f}_{d} =$frequency detected by observer
  ${v}_{s} =$speed of sound wave
  ${v}_{o} =$speed of object
  $f =$frequency at source
  
  ${f}_{d} = \left(\frac{\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{340 \frac{m}{s}}}{\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{340 \frac{m}{s}} - 25 \frac{m}{s}}\right) 300.0 H z$
  
  $\approx \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{320 H z} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
  
  Since the detected frequency is higher than $300.0 H z$, the frequency at the source, we can conclude that our answer is correct.