Unit 7: Oscillations
Weightage : 10-15%
Pretend that someone managed to dig a hole straight through the center of the Earth all the way to the other side. If an object were dropped into that hole, which of the following would best describe its motion?
Assume ideal conditions: the Earth is a perfect sphere, there are no dissipative forces, and the object cannot be destroyed.
Correct Answer: \( \boxed{\mathrm{C}} \)
Inside a uniform spherical Earth, the gravitational force on an object is directly proportional to its distance from the center:
\( F=-kr \)
where \(r\) is the displacement from the Earth’s center. This restoring force has the same form as Hooke’s law for a spring, so the object undergoes simple harmonic motion.
The object starts from rest with maximum gravitational potential energy. As it falls toward the center, its potential energy is converted into kinetic energy, reaching maximum speed at the center.
It then continues toward the opposite side, where its kinetic energy is converted back into potential energy. Since there are no dissipative forces, the total mechanical energy remains constant.
Therefore, the object continues oscillating between the two openings with constant amplitude and period.
Hence, the correct answer is (C).
A ball is dropped from a height of \(10~\mathrm{m}\) onto a hard surface. Assume the collision with the surface is perfectly elastic.
Under these conditions, the motion of the ball is
Correct Answer: \( \boxed{\mathrm{D}} \)
The time required for the ball to fall \(10~\mathrm{m}\) is
\(t=\sqrt{\dfrac{2h}{g}}=\sqrt{\dfrac{2(10)}{9.8}}\approx1.43~\mathrm{s}\)
Since the collision with the ground is perfectly elastic, the ball rebounds to its original height in the same amount of time.
Therefore, the period of the motion is
\(T=2t\approx2(1.43)\approx2.86~\mathrm{s}\approx2.8~\mathrm{s}\)
Although the motion repeats with a constant period, it is not simple harmonic motion because the restoring force is not proportional to the displacement from an equilibrium position. Instead, the ball undergoes free fall under the constant gravitational force.
Therefore, the motion is periodic with a period of about \(2.8~\mathrm{s}\), but it is not simple harmonic motion. Hence, the correct answer is (D).
An object is initially hanging in equilibrium from a vertical spring. The object is pulled down \(15\ \mathrm{cm}\) from its equilibrium position, as illustrated above, and released at time \(t=0\). The object then oscillates with a period of \(2.0\ \mathrm{s}\). Let \(x=0\) be the equilibrium position and let the positive direction be upward. What is the position of the object at \(t=3.0\ \mathrm{s}\)?
Correct Answer: \( \boxed{\mathrm{A}} \)
The amplitude of the motion is
\( A = 15\ \mathrm{cm} \)
Since the object is pulled downward and released from rest, its initial position is
\( x(0)=-15\ \mathrm{cm} \)
The period of oscillation is
\( T=2.0\ \mathrm{s} \)
After one full period (\(2.0\ \mathrm{s}\)), the object returns to the same position:
\( x(2.0\ \mathrm{s})=-15\ \mathrm{cm} \)
The additional \(1.0\ \mathrm{s}\) is equal to
\( \frac{T}{2} \)
After half a period, the object is at the opposite extreme of its motion:
\( x(3.0\ \mathrm{s})=+15\ \mathrm{cm} \)
Therefore, the position of the object at \(t=3.0\ \mathrm{s}\) is
\( \boxed{+15\ \mathrm{cm}} \)
Thus, the correct answer is (A).
