A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered by the observer is f1. If the speed of the train is reduced to 17 m/s, the frequency registered is f2. If speed of sound is 340 m/s, then the ratio \(\frac{{{f_1}}}{{{f_2}}}\) is
1. \(\frac{{19}}{{18}}\)
2. \(\frac{{21}}{{20}}\)
3. \(\frac{{20}}{{19}}\)
4. \(\frac{{18}}{{17}}\)
1. \(\frac{{19}}{{18}}\)
2. \(\frac{{21}}{{20}}\)
3. \(\frac{{20}}{{19}}\)
4. \(\frac{{18}}{{17}}\)
Correct Answer – Option 1 : \(\frac{{19}}{{18}}\)
Concept:
When a source is moving towards a stationary observer, observed frequency is given by
\({f_{observed}} = f\left( {\frac{v}{{v + {v_s}}}} \right)\)
Where, f = frequency of sound from the source,
v = speed of sound and
vs = speed of source.
Calculation:
Given,
Speed of the train = 34 m/s
Frequency when the whistle blows is f1
Speed of the train is reduced to 17 m/s after sometime
Speed of the sound = 340 m/s
Frequency registered after = f2
Now applying above formula to two different conditions given in problem, we get
f1 = Observed frequency initially
\(\Rightarrow {f_1} = f\left( {\frac{{340}}{{340 – 34}}} \right)\)
\(\therefore {f_1} = f\left( {\frac{{340}}{{306}}} \right)\)
f2 = Observed frequency when speed of source is reduced
\(\Rightarrow {f_2} = f\left( {\frac{{340}}{{340 – 17}}} \right)\)
\(\therefore {f_2} = f\left( {\frac{{340}}{{323}}} \right)\)
So, the ratio f1 : f2 is
\(\Rightarrow \frac{{{f_1}}}{{{f_2}}} = \frac{{f\left( {\frac{{340}}{{306}}} \right)}}{{f\left( {\frac{{340}}{{323}}} \right)}} = \frac{{340}}{{306}} \times \frac{{323}}{{340}}\)
\(\Rightarrow \frac{{{f_1}}}{{{f_2}}} = \frac{{323}}{{306}} = \frac{{19}}{{18}}\)
\(\therefore \frac{{{f_1}}}{{{f_2}}} = \frac{{19}}{{18}}\)
Therefore, then the ratio \(\frac{{{f_1}}}{{{f_2}}}{\rm{\;is\;given\;by}}\:\frac{{19}}{{18}}\).