Question

In a resonance column, first and second resonance are obtained at depths $22.7cm$ and $70.2cm.$ The third resonance will be obtained at a depth:
A.$117.7cm$
B.$92.9cm$
C.$115.5cm$
D.$113.5cm$

Answer 1

Hint: The Resonance is indicated by the sudden spike in the intensity of the sound when the column is adjusted to the optimum length. The resonance is a standing wave phenomenon in the air column and happens when the column length is: λ/4, 3λ/4, 5λ /4
where λ is the sound wavelength.

Complete step by step solution:
Let the three resonating points are ${{R}_{1}},{{R}_{2}}$and ${{R}_{3}}$ at positions ${{y}_{1}},{{y}_{2}}$ and ${{y}_{3}}$from the reference point `O’.
Given that
${{y}_{1}}=22.7cm$
${{y}_{2}}=70.2cm$
So the position or depth of the third resonating point is
${{y}_{3}}={{y}_{2}}+RL.......\left( i \right)$
Where RL is resonance length, here remains that the separation between the two consecutive resonating points is known as resonance length (RL).
$RL={{y}_{2}}-{{y}_{1}}$
$RL=70.2-22.7$
$RL=47.5cm$
So from equation (i) the position or depth of the third resonating point is given as,
${{y}_{3}}={{y}_{2}}+RL$
${{y}_{3}}=70.2+47.5$
${{y}_{3}}=117.7cm$

Hence, option (A) is correct.

Additional Information: Resonance only occurs when the first object is vibrating at the natural frequency of the second object when the match is achieved, the tuning fork forces the air column inside of the resonance tube to vibrate at its natural frequency and resonance is achieved, this is about first resonance $\left( {{R}_{1}} \right).$ At the natural frequency of a part is usually referred to as the first resonance in the same part and again occurs at a higher frequency, with next highest frequency. This is about second resonances $\left( {{R}_{2}} \right)$ the next the third, and so on.

Note: Resonance discusses the phenomenon of increased amplitude that occurs when the frequency of a period ally applied force is equal to or close to a natural frequency of the system on which it acts.