Simple Harmonic Motion 5 Question 4

4. A cylindrical plastic bottle of negligible mass is filled with 310 $mL$ of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency $\omega$. If the radius of the bottle is $2.5 cm$, then $\omega$ is close to (Take, density of water $=10^{3} kg / m^{3}$ )

(a) $2.50 rad s^{-1}$

(b) $5.00 rad s^{-1}$

(c) $1.25 rad s^{-1}$

(d) $3.75 rad s^{-1}$

(2019 Main, 10 Jan II)

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Answer:

Correct Answer: 4. $(*)$

Solution:

  1. In equilibrium condition bottle floats in water and its length ’ $l$ ’ inside water is same as the height of water upto which bottle is filled.

So, $l=$ Volume of water in bottle/Area

$$ =\frac{310}{\pi \times(2.5)^{2}}=15.8 cm=0.158 m $$

When bottle is slightly pushed inside by an amount $x$ then, restoring force acting on the bottle is the upthrust of fluid displaced when bottle goes into liquid by amount $x$.

So, restoring force is;

$$ F=-(\rho A x) g $$

where $\rho=$ density of water,

$A=$ area of cross-section of bottle and

$x=$ displacement from equilibriumposition

But $F=m a$

where, $\quad m=$ mass of water and bottle system

$$ =A l \rho $$

From (i) and (ii) we have,

$$ A l \rho \alpha=-\rho A x g \text { or } a=-\frac{g}{l} x $$

As for SHM, $a=-\omega^{2} x$

We have $\omega=\sqrt{\frac{g}{l}}=\sqrt{\frac{10}{0.158}}=\sqrt{63.29} \approx 8 rad s^{-1}$

$\therefore$ No option is correct.

5 We know that in case of torsonal oscillation frequency

$$ v=\frac{k}{\sqrt{I}} $$

where, $I$ is moment of inertia and $k$ is torsional constant.

$\therefore$ According to question, $v _1=\frac{k}{\sqrt{\frac{M(2 L)^{2}}{12}}}$

(As, MOI of a bar is $I=\frac{M L^{2}}{12}$ )

or

$$ v _1=\frac{k}{\sqrt{\frac{M L^{2}}{3}}} $$

When two masses are attached at ends of rod. Then its moment of inertia is

$$ \frac{M(2 L)^{2}}{12}+2 m \frac{L}{2}^{2} $$

So, new frequency of oscillations is,

$$ \begin{aligned} v _2 & =\frac{k}{\sqrt{\frac{M(2 L)^{2}}{12}+2 m \frac{L^{2}}{2}}} \\ v _2= & \frac{k}{\sqrt{\frac{M L^{2}}{3}+\frac{m L^{2}}{2}}} \end{aligned} $$

As, $\quad v _2=80 %$ of $v _1=0.8 v _1$

So, $\quad \frac{k}{\sqrt{\frac{M L^{2}}{3}+\frac{m L^{2}}{2}}}=\frac{0.8 \times k}{\sqrt{\frac{M L^{2}}{3}}}$

After solving it, we get,

$$ \frac{m}{M}=0.37 $$



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