Question

In a resonance tube experiment, the first two resonances are observed at length $10.5cm$ and $29.5cm$. The third resonance is observed at the length
(A) $47.5cm$
(B) $58.5cm$
(C) $48.5cm$
(D) $82.8cm$

Answer 1

Hint We know that for a resonance tube experiment, the difference between successive resonance is equal to half the wavelength. Hence, we will use this concept to calculate the third resonance.
i.e. \[{\lambda _n} - {\lambda _{n - 1}} = \dfrac{\lambda }{2}......(1)\]
where,
${\lambda _n}$ is length of nth resonance.
$\lambda $ is wavelength.

Complete Step by step solution
Given: length of 1st resonance = ${\lambda _1} = 10.5cm$
            Length of 2nd resonance = ${\lambda _2} = 29.5cm$

Now difference between first and second resonance is,
$\begin{array}{*{20}{c}}
  {{\lambda _2} - {\lambda _1}}& = &{\dfrac{\lambda }{2}} \\
  {29.5 - 10.5}& = &{\dfrac{\lambda }{2}} \\
  \lambda & = &{38cm}
\end{array}$
Now we have to calculate third resonance, hence using equation (1) we get
$
  {\lambda _3} - {\lambda _2} = \dfrac{\lambda }{2} \\
  {\lambda _3} = {\lambda _2} + \dfrac{\lambda }{2} \\
  {\lambda _3} = 29.5cm + 19cm \\
  {\lambda _3} = 48.5cm \\
 $
Hence the length of third resonance is $48.5cm$.

Hence option C is correct.

Note In resonance tube experiment, resonance is obtained when the first object is vibrating at the natural frequency of the second object. When this occurs, the fork forces the resonance tube to vibrate at its own frequency and the resonance is achieved