MCQ
A source of sound of frequency $90$ vibrations/ sec is approaching a stationary observer with a speed equal to $1/10$ the speed of sound. What will be the frequency heard by the observer .... $vibrations/sec$
- A$80$
- B$90 $
- ✓$100$
- D$120$
✓
Answer
Correct option: C.
$100$
c
(c) $n' = n\,\left( {\frac{v}{{v - {v_s}}}} \right) = 90\,\left( {\frac{v}{{v - \frac{v}{{10}}}}} \right)$ = $100\frac{{Vibration}}{{\sec }}$
(c) $n' = n\,\left( {\frac{v}{{v - {v_s}}}} \right) = 90\,\left( {\frac{v}{{v - \frac{v}{{10}}}}} \right)$ = $100\frac{{Vibration}}{{\sec }}$
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