The equation to solve this is the relation between speed and frequency

In this case, we don't know the value of lambda but we are not going to need it, because we are comparing the same wave
For this, we want to know if the 4m/s is higher than 0.25Hz or not to know if the change can be idenfified. Lets remember the speed of sound is 344m/s


Based on the frequency found, the change can be identified because is higher than 0.25Hz