Final Answer:
The mathematical expression for the frequency perceived by the observer in the case of a stationary observer and a moving source is (f' = f cdot left(1 + frac{v}{v_s}right)). Option b is correct.
Step-by-step explanation:
The Doppler effect describes the shift in frequency or wavelength of a wave in relation to an observer moving relative to the source of the wave. In the given scenario of a stationary observer and a moving source, the correct expression is (f' = f cdot left(1 + frac{v}{v_s}right)), where:
f' is the perceived frequency,
f is the actual frequency of the source,
v is the speed of sound, and
v_s is the speed of the source.
The positive sign in the expression indicates that the perceived frequency increases when the source is moving toward the observer. This makes sense intuitively, as the sound waves are compressed, resulting in a higher frequency.
To understand this mathematically, consider the formula in the context of a moving source. As the source approaches the observer v_s > 0 , the term frac{v}{v_s} becomes positive, leading to an increase in the perceived frequency f' > f . Conversely, if the source is moving away v_s < 0 , the term frac{v}{v_s} becomes negative, resulting in a decrease in the perceived frequency f' < f .
In summary, the correct expression (f' = f cdot left(1 + frac{v}{v_s}right)) captures the relationship between the actual frequency, speed of sound, and the speed of the moving source in the context of a stationary observer.