Question

When a train approaches a stationary observer, the apparent frequency of the whistle is n' and when the same train recedes away from the observer, the apparent frequency is n''. Then the apparent frequency n when the observer moves with the train is –

• A. $n = \frac{n' + n''}{2}$
• B. n =$\sqrt{n'n''}$
• C. n = $\frac{2n'n''}{n' + n''}$
• D. n = $\frac{2n'n''}{n'–n''}$

Answer 1

Answer: (C)

n = $\frac{2n'n''}{n' + n''}$

Answer 2

n¢ = n ….(1)

n¢¢ = n ….(2)

from (1)

\ V $\left( \frac { \mathrm { n } ^ { \prime } } { \mathrm { n } } - 1 \right) = \frac { \mathrm { n } ^ { \prime } } { \mathrm { n } } V _ { \mathrm { s } }$

\ V_s = = $\mathrm { V } \left( \frac { \mathrm { n } ^ { \prime } - \mathrm { n } } { \mathrm { n } } \right) \times \frac { \mathrm { n } } { \mathrm { n } ^ { \prime } }$ =

Put in (2)

n'' = n n¢¢ =

n² = n¢ n¢¢