A vertical tunnel is dug up through Earth passing through the centre as shown in the figure below. 6. A particle of mass m is thrown inside the tunnel. How should the density vary with radius R inside the Earth if the particle executes perfect simple harmonic motion? (a) p(R)al/R (b) p(R)= constant (c) p(R)= constant or p(R) a 1/R2 (d) p(R) a1/R2?

Question

Answer

Density Variation with Radius for Simple Harmonic Motion

When a particle is thrown inside a vertical tunnel passing through the center of the Earth, it executes simple harmonic motion. We need to determine how the density varies with radius R inside the Earth for this motion.

Understanding Simple Harmonic Motion

In simple harmonic motion, the restoring force acting on the particle is directly proportional to its displacement from the equilibrium position and is always directed towards the equilibrium position. This force can be expressed as:

F = -kx

Where F is the restoring force, x is the displacement from equilibrium, and k is the force constant.

Applying Simple Harmonic Motion to a Particle Inside the Earth

In the case of a particle inside the Earth, the restoring force acting on the particle is due to the gravitational pull towards the center of the Earth. The force is given by:

F = -mg

Where m is the mass of the particle and g is the acceleration due to gravity.

Relationship between Density and Radius

We know that the gravitational force acting on a particle is directly proportional to its mass, and the gravitational force is given by:

F = GMm/R^2

Where G is the gravitational constant, M is the mass of the Earth, m is the mass of the particle, and R is the radius of the Earth.

Comparing the two expressions for the force, we have:

-mg = GMm/R^2

Canceling out the mass of the particle, we get:

-g = GM/R^2

Dividing both sides by g, we have:

-1 = (GM/R^2)/g

Since the density (p) is given by:

p = M/V

Where V is the volume of the Earth, and the volume of a sphere is given by:

V = (4/3)πR^3

Substituting the expression for V, we get:

p = (3M)/(4πR^3)

Conclusion

From the above expression, we can see that the density (p) is inversely proportional to the cube of the radius (R). Therefore, the correct answer is:

(d) p(R) α 1/R^2

Top Courses for Physics

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams