SATHEE: Simple Harmonic Motion 5 Question 4

Simple Harmonic Motion 5 Question 4

4. A cylindrical plastic bottle of negligible mass is filled with 310 $m L$ 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 $ω$ . If the radius of the bottle is $2.5 c m$ , then $ω$ is close to (Take, density of water $= 10^{3} k g / m^{3}$ )

(a) $2.50 r a d s^{- 1}$

(b) $5.00 r a d s^{- 1}$

(d) $3.75 r a d s^{- 1}$

(2019 Main, 10 Jan II)

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Answer:

Correct Answer: 4. $(*)$

Solution:

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

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

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;

$F = - (ρ A x) g$

where $ρ =$ density of water,

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

$x =$ displacement from equilibriumposition

But $F = m a$

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

$= A l ρ$

From (i) and (ii) we have,

$A l ρ α = - ρ A x g or a = - \frac{g}{l} x$

As for SHM, $a = - ω^{2} x$

We have $ω = \sqrt{\frac{g}{l}} = \sqrt{\frac{10}{0.158}} = \sqrt{63.29} \approx 8 r a d s^{- 1}$

$∴$ 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.

$∴$ According to question, $v_{1} = \frac{k}{\sqrt{\frac{M (2 L)^{2}}{12}}}$

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

$v_{1} = \frac{k}{\sqrt{\frac{M L^{2}}{3}}}$

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,

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