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Step-by-Step Solution
Step 1: Understanding the Problem
A tunnel is dug through the Earth along a chord such that it is at a perpendicular distance of $R/2$ from the Earth’s center (where $R$ is the Earth’s radius). A particle is released inside this frictionless tunnel. We need to determine the time period of the resulting motion.
Step 2: Recognizing the Nature of Motion
When a particle moves under gravity inside the Earth, the restoring force is proportional to its displacement from the Earth’s center (assuming the Earth has uniform density or that the variation allows a linear restoring force). This setup leads to Simple Harmonic Motion (SHM). Therefore, we aim to show that the particle undergoes SHM with a certain time period.
Step 3: Expression for the Gravitational Force Inside Earth
For a particle of mass $m$ at a distance $d$ from the center of the Earth, if the Earth is assumed to have uniform density, the gravitational acceleration $g_{\text{inside}}$ varies linearly with $d$:
$$ g_{\text{inside}} = \frac{GM\,d}{R^3}, $$
where $G$ is the gravitational constant, and $M$ is the mass of the Earth.
Step 4: Component of Force along the Tunnel
Let the displacement of the particle from the midpoint of the chord (towards one end of the tunnel) be $x$. The force on the particle toward the Earth’s center can be resolved along the tunnel. If $F = mg_{\text{inside}}$ is the magnitude of the force toward the center, the component of this force along the tunnel is:
$$ F_1 = F \cos \theta = m \, g_{\text{inside}} \cos \theta. $$
With some geometry (relating $x$ to $d \cos \theta$), this becomes proportional to $x$. Hence we write:
$$ F_1 = \frac{GMm}{R^3}\, x = \frac{m\, g_s}{R}\, x, $$
where $g_s = \frac{GM}{R^2}$ is the usual gravitational acceleration at the surface of the Earth, which we typically denote as $g$.
Step 5: Deriving the Acceleration
The acceleration $a$ (or $\alpha$) along the tunnel is then:
$$ a = \frac{F_1}{m} = \frac{g_s}{R}\, x. $$
Since $g_s \equiv g$, we have:
$$ a = \frac{g}{R}\, x. $$
Step 6: Time Period of SHM
For Simple Harmonic Motion, $a = \omega^2 \, x$ where $\omega$ is the angular frequency. Comparing, we get:
$$ \omega^2 = \frac{g}{R}, $$
thus
$$ \omega = \sqrt{\frac{g}{R}}. $$
The time period $T$ is given by:
$$ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{R}{g}}. $$
Step 7: Concluding the Result
Therefore, the time period of the particle oscillating inside this chordal tunnel is:
$$ \boxed{2\pi \sqrt{\frac{R}{g}}}. $$
This matches Option 1 from the given choices.
Below is the reference image from the provided solution:
