NEET 2015
- Exam
- NEET
- Year
- 2015
- Subject
- Physics
- Chapter
- Waves and Sound
NEET 2015 · Physics · Waves and Sound
A string is stretched between fixed points separated by 75.0 cm. It...
A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencier between these two. The lowest resonant frequency for this string is:
Options
- A205 Hz
- B10.5 Hz
- C105 Hz
- D155 Hz
Correct answer
C. 105 Hz
Step-by-step solution
Two consecutive resonant frequencies for a string fixed at both ends will be
\dfrac{nv}{2l}and\dfrac{\left(n + 1\right)v}{2l}⇒ \dfrac{\left(n + 1\right)v}{2l} - \dfrac{nv}{2l} = 420 - 315\dfrac{v}{2l} = 105Hz
Which is the minimum resonant frequency.
\dfrac{nv}{2l}and\dfrac{\left(n + 1\right)v}{2l}⇒ \dfrac{\left(n + 1\right)v}{2l} - \dfrac{nv}{2l} = 420 - 315\dfrac{v}{2l} = 105Hz
Which is the minimum resonant frequency.