SHM in liquid

Classical Mechanics Level 5

A thin rod of length $L$ and area of cross section $S$ is pivoted at its lowest point P inside a stationary , homogeneous , non viscous liquid as shown in the figure.

The rod is free to rotate in a vertical plane about a horizontal axis passing through $p$ . The density of the material of the rod $D_1$ is smaller than the density of liquid $D_2$ . The rod is displaced by a small angle $θ$ from its equilibrium position and then released.

If the body performs Simple harmonic motion, calculate t's angular fequency $\omega$

Details and Assumptions

$D_1 = 200 \text {units}$

$D_2 = 400 \text {units}$

$L = 20 \text {cm}$

$g =10 \dfrac {m}{s^2}$

Also try this.

The answer is 8.66.

1 solution

Abdulmuttalib Lokhandwala
May 31, 2014

The different forces on the rod are:

Weight of the rod acting downwards = $mg$ = $SLD_1g$

Buoyant force acting upwards = $SLD_2g$

Net thrust acting on the rod upwards; $F$ = $SL(D_1-D_2)g$

Restoring torque $τ$ = $F\frac{L}{2}sinθ$

If $θ$ is in clockwise direction then $τ$ will be in anticlockwise direction

$τ$ = - $F\frac{L}{2}sinθ$

Also $τ$ = $Iα$

Where I = $\frac{ML^{2}}{3}$

Find $α$ using above equation and then find angular frequency $ω$ using differential equation of $S.H.M$

U will get $ω$ = $\sqrt{\frac{3g}{2L}\frac{D_2-D_1}{D_1}}$

It is interesting,, that in this case as D2= 2D1, so the SHM is just as if it was in air,, except,, upwards directing

Mvs Saketh - 6 years, 8 months ago

Yeah that's good observation.

rajdeep brahma - 3 years, 4 months ago

Use d_1 for $d_1$ .

Shaan Vaidya - 7 years ago

Thanks for your suggestion

abdulmuttalib lokhandwala - 7 years ago

SHM in liquid

The answer is 8.66.

1 solution

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