Doppler Effect With Source And Observer Moving

When both source and observer move along the line joining them, the general Doppler formula is $f^{\prime }=f\left(\frac{v+v_o}{v-v_s}\right)$, where $v_o$ is the observer's speed toward the source and $v_s$ is the source's speed toward the observer. The numerator grows because the observer rushes into the wavefronts, and the denominator shrinks because the source chases its own sound. Here $f=500\text{ Hz}$, $v=340\text{ m/s}$, $v_o=10\text{ m/s}$, and $v_s=20\text{ m/s}$, giving $f^{\prime }=500\left(\frac{340+10}{340-20}\right)=500\times \frac{350}{320}=500\times 1.094\approx 547\text{ Hz}$. The value $515\text{ Hz}$ accounts for only the observer's motion and omits the source term. The value $530\text{ Hz}$ uses only the source motion. The value $485\text{ Hz}$ wrongly treats the trains as receding, lowering the pitch. This is the NCERT combined Doppler relation. As a plausibility check, both motions act to raise the pitch since the trains close in, so the result must exceed 500 Hz and be larger than either single-effect value, which the 547 Hz answer correctly reflects.