Question

A small heavy metallic sphere of specific gravity 2 is suspended by a light inextensible string has a period of oscillation t .now it is immersed in water as shown ib the figure . its period of oscillation now is

Answer 1

Answer:

New time period of the given metallic sphere in water is

$T' = \sqrt2 T$

Explanation:

When heavy metallic sphere is suspended in air then restoring force on it while it is displaced by small angle

$F_r = -mg \theta$

$a = \frac{g}{L}x$

so time period is given as

$T = 2\pi\sqrt{\frac{L}{g}}$

Now when we immerse the sphere into the water then we have

$F_{net} = mg - F_b$

$F_{net} = mg - \rho_w Vg$

$F_{net} = mg(1 - \frac{\rho_w}{\rho_o})$

so we have

$a = - \frac{g(1 - \frac{\rho_w}{\rho_o})}{L}x$

so new time period is given as

$T' = 2\pi\sqrt{\frac{L}{g(1 - \frac{\rho_w}{\rho_o})}}$

$\rho_o = 2\rho_w$

so we have

$T' = 2\pi\sqrt{\frac{2L}{g}}$

so we have

$T' = \sqrt2 T$

#Learn

Topic : Simple pendulum

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