Problem 10 - University Physics with Modern Physics

Question

14.10 A pendulum is suspended from the ceiling of an elevator. When the elevator is at rest, the period of the pendulum is \(T\). The elevator accelerates upward, and the period of the pendulum is then a) still T. b) less than \(T_{-}\) c) greater than \(T\).

Step-by-Step Solution

Answer

Question: When an elevator accelerates upward, the period of a pendulum suspended from the ceiling of the elevator becomes: a) equal to the original period \(T\) b) less than the original period \(T\) c) greater than the original period \(T\) Answer: b) less than the original period \(T\)

Step 1: State the formula for the period of a pendulum

The formula for the period of a pendulum is given by: \[T = 2\pi\sqrt{\frac{l}{g}}\] where \(T\) is the period, \(l\) is the length of the pendulum, and \(g\) is the acceleration due to gravity.

Step 2: Analyze the effect of the elevator's upward acceleration on the pendulum

When the elevator accelerates upward, it creates an additional force acting on the pendulum, in the upward direction. This force increases the effective gravity acting on the pendulum. Let's represent the upward acceleration as \(a\). Since the acceleration is acting in the same direction as gravity, the effective gravity acting on the pendulum can be written as: \[g' = g + a\]

Step 3: Find the period of the pendulum with the new effective gravity

Since the new effective gravity is \(g'\), the formula for the period of the pendulum when the elevator is accelerating upward can be written as: \[T' = 2\pi\sqrt{\frac{l}{g'}}\] Substitute the value of \(g'\) from step 2: \[T' = 2\pi\sqrt{\frac{l}{g + a}}\]

Step 4: Compare the new period \(T'\) with the original period \(T\)

We can compare the two periods by finding the ratio between them: \[\frac{T'}{T} = \frac{2\pi\sqrt{\frac{l}{g + a}}}{2\pi\sqrt{\frac{l}{g}}}\] The \(2\pi\sqrt{\frac{l}{g}}\) terms cancel out: \[\frac{T'}{T} = \sqrt{\frac{g}{g+a}}\] Since \(g > 0\) and \(a > 0\), the denominator \((g+a)\) will always be greater than \(g\). Therefore, as the denominator increases while the numerator remains constant, the ratio \(\frac{g}{g + a}\) will be less than 1. Thus, the period \(T'\) of the pendulum when the elevator is accelerating upward will be less than the original period \(T\).

Step 5: State the answer

Based on our analysis, the period of the pendulum \(T'\) when the elevator is accelerating upward will be less than the original period \(T\). Hence, the correct answer is: (b) less than \(T\).

Key Concepts

Simple Harmonic Motion

Simple harmonic motion is a type of repetitive movement, often found in pendulums or springs. It's characterized by an oscillating object moving back and forth around an equilibrium position. A great example is a pendulum, which swings from side to side. One of the key features of this kind of motion is that the restoring force, or "push" that brings the object back to the center, is always proportional to the displacement. This means the farther away the pendulum is from the center, the stronger the force acting on it to bring it back.

For pendulums:

• The period, or the time it takes for one complete swing (back and forth), is constant and depends on the length of the pendulum and the local acceleration due to gravity.
• The formula for the period is:
  • \(T = 2\pi\sqrt{\frac{l}{g}} \)

This shows that changes in either the pendulum's length or the gravity will impact the period of oscillation. That's why understanding variables like effective gravity is crucial.

Effective Gravity

Effective gravity is the apparent change in the gravitational pull experienced by an object due to external accelerations. When in an elevator, for instance, you might feel heavier or lighter as it moves up or down. This change in perceived weight is due to effective gravity.

In terms of the pendulum, effective gravity can alter the period of oscillation:

• When an elevator moves downward, the effective gravity decreases as it subtracts from the actual gravitational acceleration.
• Conversely, when moving upward, it adds to the gravitational pull, increasing effective gravity.

Considering the formula,

• The adjusted effective gravity becomes: \(g' = g + a\)

Where \(a\) is the elevator's acceleration. This modification affects the pendulum's frequency of motion by changing the acceleration due to gravity that it "feels".

Elevator Acceleration

Elevator acceleration can significantly affect the physics of objects within it, altering experiences of weight and motion. As the elevator accelerates upward, anyone inside might feel a stronger force pressing them against the floor. For pendulums, this upward acceleration effectively boosts gravity, as the elevator is pushing more upward forces in addition to gravity.

Let's break it down:

• The acceleration \(a\) plays a crucial role when the elevator moves.
• When moving upward, the acceleration adds to the gravitational pull experienced by objects, acting like an additional force of gravity.

Thus, a pendulum swinging inside experiences a stronger gravitational acceleration than it would if the elevator were still.

This leads to a decrease in the pendulum's period, as given by the formula:

• \(T' = 2\pi\sqrt{\frac{l}{g + a}}\)

With the addition of \(a\), the pendulum swings more quickly compared to when it's in a stationary elevator.