Question

In the following four
(i) The time period of revolution of a satellite just above the earth’s surface $\left( {{T_{st}}} \right)$
(ii) The time period of oscillation of ball inside the tunnel bored along the diameter of the earth $\left( {{T_{ma}}} \right)$
(iii) The time period of a simple pendulum having a length equal to the earth’s radius in a uniform field of 9.8 newton/kg $\left( {{T_{sp}}} \right)$
(iv) The time period of an infinite length simple pendulum in the earth’s gravitational field $\left( {{T_{is}}} \right)$

Which of the following is true
a. ${T_{st}} > {T_{ma}}$
b. ${T_{st}} < {T_{ma}}$
c. ${T_{sp}} > {T_{is}}$
d. All the time periods are equal.

Answer 1

The time period is defined as the time taken by a complete cycle of revolving or oscillating objects to pass a point. The time period is affected by gravity therefore the time period in the above conditions may differ. In the first statement, the satellite is revolving just above the earth’s surface. In this case, we will consider the radius of the orbit equal to the radius of the earth. Use the time period formula in every case to compare them.

Complete step by step answer:
Step 1: The time period of the revolution of a satellite just above the earth’s surface can be calculated by taking the radius of orbit equal to the radius of the earth. Therefore the time period of the satellite is given by
${T_{st}} = \dfrac{{2\pi R}}{v}$
But we know that the orbital velocity of a revolving object can be written as $v = \sqrt {\dfrac{{GM}}{R}}$ ,
$\therefore {T_{st}} = 2\pi R\sqrt {\dfrac{R}{{GM}}}$
$\Rightarrow {T_{st}} = 2\pi \dfrac{{{R^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}}}{{\sqrt {GM} }}$ …….equation (1)
We know that the acceleration due to the gravity of a satellite is given by $g = \dfrac{{GM}}{{{R^2}}}$ . Put this value in the first equation,
$\therefore {T_{st}} = 2\pi \sqrt {\dfrac{R}{g}}$

Step 2: Now calculate the time period of oscillation of the ball inside the tunnel bored along the diameter of the earth $\left( {{T_{ma}}} \right)$ . First express the relation among the angular speed, amplitude, and acceleration for a simple harmonic motion.
$\therefore a = A{\omega ^2}$ , where $a$ is the acceleration, $A$ is the amplitude, and $\omega$ is the angular speed.
Here, as the body reaches the end of the tunnel the acceleration becomes equal to the earth’s acceleration and the amplitude becomes equal to the radius of the earth.
$\therefore \omega = \sqrt {\dfrac{g}{R}}$ …..equation (2)
But we have the other expression for the angular speed $\omega$ in terms of the time period
$\therefore \omega = \dfrac{{2\pi }}{{{T_{ma}}}}$ , where ${T_{ma}}$ is the time period.
Substituting the value of $\omega$ into the second equation
$\therefore \dfrac{{2\pi }}{{{T_{ma}}}} = \sqrt {\dfrac{g}{R}}$
${T_{ma}} = 2\pi \sqrt {\dfrac{R}{g}}$
Step 3: Calculate the time period of a simple pendulum having a length equal to the earth’s radius in a uniform field of 9.8 newton/kg $\left( {{T_{is}}} \right)$ . Write the expression for the time period of the pendulum moving in simple harmonic motion
$\therefore {T_{sp}} = 2\pi \sqrt {\dfrac{1}{{g\left( {\dfrac{1}{l} + \dfrac{1}{R}} \right)}}}$ , where $l$ is the length of the simple pendulum.
But in the question the length of the pendulum is equal to the radius of the earth, therefore
$\therefore {T_{sp}} = 2\pi \sqrt {\dfrac{1}{{g\left( {\dfrac{1}{R} + \dfrac{1}{R}} \right)}}}$
$\Rightarrow {T_{sp}} = 2\pi \sqrt {\dfrac{R}{{2g}}}$

Step 4: Now let us calculate the time period of an infinite length simple pendulum in the earth’s gravitational field $\left( {{T_{is}}} \right)$ . The formula for the time period will be the same as in step 3. The only difference is that the length of the pendulum is infinity. Therefore,
$\therefore {T_{is}} = 2\pi \sqrt {\dfrac{1}{{g\left( {\dfrac{1}{\infty } + \dfrac{1}{R}} \right)}}}$
We know that the value of $\dfrac{1}{\infty }$ is zero
$\therefore {T_{is}} = 2\pi \sqrt {\dfrac{1}{{g\left( {\dfrac{1}{R}} \right)}}}$
$\Rightarrow {T_{is}} = 2\pi \sqrt {\dfrac{R}{g}}$

Step 5: Now consider option A. Since ${T_{st}}$ and ${T_{ma}}$ both are equal therefore option A is incorrect. Option B is also incorrect for the same reason.

Step 6: Consider option C. Clearly, ${T_{sp}} > {T_{is}}$ therefore the option C is correct.

Step 7: Option D cannot be correct as ${T_{sp}}$ is not equal to ${T_{is}}$.

Hence, the correct answer is option (C).

Note: Oscillation is a harmonic motion in which the object vibrates. The motion is called harmonic because musical instruments make such vibrations that in turn cause corresponding sound waves in the air. Therefore the oscillation should not be confused with another type of motion.