Answer:
The observed wavelength decreases
Explanation:
When a source of a wave is moving relative to an observer, there is an apparent change in the frequency perceived by the observer (this is called Doppler effect), and the apparent frequency is given by
[tex]f'=(\frac{v\pm v_o}{v+v_s})f[/tex]
where
f is the original frequency
f' is the apparent frequency
[tex]v_o[/tex] is the velocity of the observer, and it is positive if the observer is moving towards the source and negative if it is moving away
[tex]v_s[/tex] is the velocity of the source, and it is positive if the source is moving away from the observer and negative if it is approaching it
v is the speed of the waves
In this case, the object is approaching the observer (which can be considered at rest): this means that the fraction [tex]\frac{v\pm v_o}{v+v_s}[/tex] is greater than 1 (because [tex]v_o = 0[/tex] while [tex]v_s[/tex] is negative), so
[tex]f>f'[/tex]
which means that the apparent frequency is higher than the real frequency. And since the wavelength is inversely proportional to the frequency, according to the equation:
[tex]\lambda = \frac{v}{f}[/tex]
it means that the apparent wavelength will be shorter than the real wavelength.
