Question

What conclusion can be obtained from the observation that when the prongs of a sound making tuning fork touch a beaker, the water gets splashed?
(A) That the prongs of the tuning fork are vibrating.
(B) The prongs of the tuning fork produce ultrasound.
(C) Waves are produced on the surface of water.
(D) All the above.

Answer 1

Hint
We know that any object moving in Simple Harmonic Motion produces areas of compressions and rarefactions. We can make this analogical to our problem.
Formulae Used: $ y = Asin\left( {\omega t + \phi } \right) $
Where, $ y $ is the displacement of the oscillating body at any time $ t $, $ A $ is the amplitude, $ \omega $ is the angular velocity and $ \phi $ is the initial phase or epoch.

Complete step by step answer
First, let us calculate the total energy of the oscillating body.
Clearly, there are two type of energies contributing to the total energy of the body,
 $\Rightarrow E = {E_k} + {E_p} $
One is the Kinetic Energy, $ {E_k} = {\text{ }}\raise.5ex\hbox{ $ \scriptstyle 1 $ }\kern-.1em/
 \kern-.15em\lower.25ex\hbox{ $ \scriptstyle 2 $ } {\text{ }}m{v^2} $, where $ m $ is the mass of the mass of the body and $ v $ is its velocity.
And, the other is the Potential Energy, $ {E_p} = {\text{ }}\int_0^y {kydy} $, where $ k $ is the force or spring constant, $ y $ is the displacement of the body at any time $ t $ .
Now, if $ y = asin\left( {\omega t + \phi } \right) $
And, we know $ v = dy/dt $
Thus, $ v = {\text{ }}d\left[ {asin\left( {\omega t + \phi } \right)} \right]/dt $
After differentiation, we get
 $\Rightarrow v = {\text{ }}a\omega cos\left( {\omega t + \phi } \right) $
Now,We get,
 $\Rightarrow {E_k} = {\text{ }}\raise.5ex\hbox{ $ \scriptstyle 1 $ }\kern-.1em/
 \kern-.15em\lower.25ex\hbox{ $ \scriptstyle 2 $ } {\text{ }}m{\left[ {A\omega cos\left( {\omega t + \phi } \right)} \right]^2} $
 $\Rightarrow \Rightarrow {E_k} = {\text{ }}\raise.5ex\hbox{ $ \scriptstyle 1 $ }\kern-.1em/
 \kern-.15em\lower.25ex\hbox{ $ \scriptstyle 2 $ } {\text{ }}m{A^2}{\omega ^2}co{s^2}\left( {\omega t + \phi } \right) \cdot \cdot \cdot \cdot (1) $
Now, we know,
From Newton’s Second Law, we get
 $\Rightarrow F = m{\omega ^2}y $
Also, $ F = - ky $
But the negative symbol only tells us about the direction of the force.
So, we can take only the magnitude for our problem.
So comparing, we get
 $\Rightarrow m{\omega ^2}y = ky $
 $\Rightarrow k = m{\omega ^2} $
Now,Acceleration,
 $\Rightarrow a = {d^2}y/d{t^2} $
 $\Rightarrow a = - A{\omega ^2}sin\left( {\omega t + \phi } \right) = - {\omega ^2}y $
Also, $ {E_p} = \int_0^y {kydy} $
Thus after integration, we get
 $\Rightarrow {E_p} = {\text{ }}\raise.5ex\hbox{ $ \scriptstyle 1 $ }\kern-.1em/
 \kern-.15em\lower.25ex\hbox{ $ \scriptstyle 2 $ } {\text{ }}k{y^2} $
Thus, $ E = {\text{ }}\raise.5ex\hbox{ $ \scriptstyle 1 $ }\kern-.1em/
 \kern-.15em\lower.25ex\hbox{ $ \scriptstyle 2 $ } {\text{ }}m{A^2}{\omega ^2}co{s^2}\left( {\omega t + \phi } \right){\text{ }} + {\text{ }}\raise.5ex\hbox{ $ \scriptstyle 1 $ }\kern-.1em/
 \kern-.15em\lower.25ex\hbox{ $ \scriptstyle 2 $ } {\text{ }}k{y^2} $
After substituting all the values and then evaluating, we get
 $\Rightarrow E = {\text{ }}\raise.5ex\hbox{ $ \scriptstyle 1 $ }\kern-.1em/
 \kern-.15em\lower.25ex\hbox{ $ \scriptstyle 2 $ } {\text{ }}m{A^2}{\omega ^2} $
Now,Substituting $ k = m{\omega ^2} $, we get
 $\Rightarrow E = {\text{ }}\raise.5ex\hbox{ $ \scriptstyle 1 $ }\kern-.1em/
 \kern-.15em\lower.25ex\hbox{ $ \scriptstyle 2 $ } {\text{ }}k{A^2} $
This is the energy which is transmitted to the surroundings by an oscillating body producing regions of compressions and rarefactions which is apparent here by the splashing of water.
Thus, the correct option is (A).

Note
While calculating the potential energy, we took the limit of integration as $ 0{\text{ }}and{\text{ }}y $. It is also possible to take $ {y_1} and {\text{ }}\left( {{y_1} + y} \right) $. Just the difference should be $ y $ as it becomes easier to calculate.