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Simple Harmonic Motion question

2004 · Shift 0 · Q176

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Simple Harmonic Motion
2004 · Shift 0 · Q176

Simple Harmonic Motion question

2004 · Shift 0 · Q176

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1

The bob of a simple pendulum executes simple harmonic motion in water with a period

t,

while the period of oscillation of the bob is

{t_0}

in air. Neglecting frictional force of water and given that the density of the bob is

\left( {4/3} \right) \times 1000\,\,kg/{m^3}.

What relationship between

t

and

{t_0}

is true

A
$t = 2{t_0}$
B
$t = {t_0}/2$
C
$t = {t_0}$
D
$t = 4{t_0}$

View written solutionFree

Correct answer: A

Time period of a simple pendulum in air

For small oscillations, the time period in air is

t_0 = 2\pi \sqrt{\frac{l}{g}}.

Effect of immersing the bob in water

When the bob is fully immersed in water, it experiences an upward buoyant force.

Weight of bob: $W = mg = \rho_b V g$
Buoyant force: $B = \rho_w V g$

So the effective weight becomes

W_{\text{eff}} = W - B = (\rho_b - \rho_w)Vg.

Hence the effective acceleration due to gravity is

g_{\text{eff}} = g\left(1 - \frac{\rho_w}{\rho_b}\right).

Use the given density

Given density of bob:

\rho_b = \left(\frac{4}{3}\right)\times 1000\; \text{kg/m}^3

Density of water:

\rho_w = 1000\; \text{kg/m}^3

Therefore,

\frac{\rho_w}{\rho_b} = \frac{1000}{(4/3)1000} = \frac{3}{4}.

So,

g_{\text{eff}} = g\left(1 - \frac{3}{4}\right) = \frac{g}{4}.

Time period in water

Now the period in water is

t = 2\pi \sqrt{\frac{l}{g_{\text{eff}}}} = 2\pi \sqrt{\frac{l}{g/4}}.

Thus,

t = 2\pi \sqrt{\frac{4l}{g}} = 2\, 2\pi \sqrt{\frac{l}{g}} = 2t_0.

Check options

A: $t = 2t_0$ ✅
B: $t = t_0/2$ ❌
C: $t = t_0$ ❌
D: $t = 4t_0$ ❌

Therefore, the correct option is A.

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