A simple pendulum doing small oscillations at a place R height above earth surface has time period of T_1=4 ~s . T_2 would be it's time period if it is brought to a point which is at a height 2 R from earth surface. Choose the correct relation [ R= radius of earth] :

Simple Harmonic Motion PYQ — JEE Main 2024 · 5 Apr · Shift 1 · Q66 with Solution

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Correct answer: A

Time period of a simple pendulum

For small oscillations,

T = 2\pi \sqrt{\frac{l}{g}}

where $l$ is the length of pendulum and $g$ is acceleration due to gravity at that place.

So, for the same pendulum,

T \propto \frac{1}{\sqrt{g}}

Gravity at height $h$ above earth

Acceleration due to gravity at height $h$ is

g_h = g\left(\frac{R}{R+h}\right)^2

where $R$ is radius of earth.

At height $R$ above earth surface

Here $h=R$ . So,

g_1 = g\left(\frac{R}{R+R}\right)^2 = g\left(\frac{R}{2R}\right)^2 = \frac{g}{4}

Given time period here is

T_1 = 4\,\text{s}

At height $2R$ above earth surface

Here $h=2R$ . So,

g_2 = g\left(\frac{R}{R+2R}\right)^2 = g\left(\frac{R}{3R}\right)^2 = \frac{g}{9}

Relate $T_1$ and $T_2$

Since

T \propto \frac{1}{\sqrt{g}}

we have

\frac{T_2}{T_1} = \sqrt{\frac{g_1}{g_2}}

Substitute $g_1=\frac{g}{4}$ and $g_2=\frac{g}{9}$ :

\frac{T_2}{T_1} = \sqrt{\frac{g/4}{g/9}} = \sqrt{\frac{9}{4}} = \frac{3}{2}

Therefore,

T_2 = \frac{3}{2}T_1

Rearranging,

3T_1 = 2T_2

Since $T_1=4\,\text{s}$ ,

T_2 = \frac{3}{2}\times 4 = 6\,\text{s}

Check options

A: $3T_1=2T_2$ ✅ correct
B: $T_1=T_2$ ❌
C: $2T_1=3T_2$ ❌
D: $2T_1=T_2$ ❌

Hence, the correct option is A.

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View written solutionFree

Correct answer: A

Time period of a simple pendulum

For small oscillations,

T = 2\pi \sqrt{\frac{l}{g}}

where $l$ is the length of pendulum and $g$ is acceleration due to gravity at that place.

So, for the same pendulum,

T \propto \frac{1}{\sqrt{g}}

Gravity at height $h$ above earth

Acceleration due to gravity at height $h$ is

g_h = g\left(\frac{R}{R+h}\right)^2

where $R$ is radius of earth.

At height $R$ above earth surface

Here $h=R$ . So,

g_1 = g\left(\frac{R}{R+R}\right)^2 = g\left(\frac{R}{2R}\right)^2 = \frac{g}{4}

Given time period here is

T_1 = 4\,\text{s}

At height $2R$ above earth surface

Here $h=2R$ . So,

g_2 = g\left(\frac{R}{R+2R}\right)^2 = g\left(\frac{R}{3R}\right)^2 = \frac{g}{9}

Relate $T_1$ and $T_2$

Since

T \propto \frac{1}{\sqrt{g}}

we have

\frac{T_2}{T_1} = \sqrt{\frac{g_1}{g_2}}

Substitute $g_1=\frac{g}{4}$ and $g_2=\frac{g}{9}$ :

\frac{T_2}{T_1} = \sqrt{\frac{g/4}{g/9}} = \sqrt{\frac{9}{4}} = \frac{3}{2}

Therefore,

T_2 = \frac{3}{2}T_1

Rearranging,

3T_1 = 2T_2

Since $T_1=4\,\text{s}$ ,

T_2 = \frac{3}{2}\times 4 = 6\,\text{s}

Check options

A: $3T_1=2T_2$ ✅ correct
B: $T_1=T_2$ ❌
C: $2T_1=3T_2$ ❌
D: $2T_1=T_2$ ❌

Hence, the correct option is A.

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