First slide

Applications of SHM

Question

A cylindrical plastic bottle of negligible mass is filled with 500ml of water. On slightly pressing it downwards and releasing, it starts executing simple harmonic motion. If radius of the bottle is 5cm, then its angular frequency of oscillation is close to ( $d e n s i t y o f w a t e r ρ = 10^{3} k g / m^{3}$ )

Difficult

A

3.9 rad/s
B

12.7 rad/s
C

153 rad/s
D

127 rad/s

Solution

The restoring force responsible for executing simple harmonic motion in this case is the buoyant force.
Using this, we can equate $F_{B} = ρ V g$ to the restoring force $F = k x$ .
Comparing the two equations, $k x = ρ V g$
Here $V = A x$ ,
Substituting this we get $k x = ρ A x g$
Hence, $k = ρ A g$
For a body executing simple harmonic motion, its angular frequency can be written as    $ω = \sqrt{\frac{k}{m}}$
Substituting $k = ρ A g$ in $ω = \sqrt{\frac{k}{m}}$ , we get
$ω = \sqrt{\frac{ρ A g}{m}}$ .
Now, the mass of water in the bottle $m = ρ A h$
Hence,    $ω = \sqrt{\frac{ρ A g}{ρ A h}} = \sqrt{\frac{g}{h}}$ ………………………(1)
In the question, the volume of water is given as 500 ml. Using the conversion that $1 l = 10^{- 3} m^{3},$ we get $1 m l = 10^{- 6} m^{3},$
Hence, volume of 500ml of water is    $5 \times 10^{- 4} m^{3}$
Again using $V = A h$ , $5 \times 10^{- 4} = A h$
Therefore $h = \frac{5 \times 10^{- 4}}{A} = \frac{5 \times 10^{- 4}}{π r^{2}}$ . Given the radius if the base is 5 cm,
$h = \frac{5 \times 10^{- 4}}{π \times {(5 \times 10^{- 2})}^{2}}$ = $\frac{1}{5 π}$
$h = \frac{1}{15.70}$ .
Substituting this in equation (1),
$ω = \sqrt{\frac{g}{h}} = \sqrt{\frac{9.8}{\frac{1}{15.70}}}$
$ω = 12.7 r a d / s$

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