Problem 9
Question
If you double the length of a pendulum, what happens to its period?
Step-by-Step Solution
Verified Answer
When the pendulum length is doubled, its period increases by a factor of \( \sqrt{2} \) or about 1.414.
1Step 1: Understand the Pendulum Period Formula
The period of a pendulum, which is the time it takes to complete one oscillation, is given by the formula: \[ T = 2\pi \sqrt{\frac{L}{g}} \]where \( T \) is the period, \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity. This equation shows that the period depends on the square root of the pendulum's length.
2Step 2: Analyze the Effect of Doubling the Length
When the length \( L \) of the pendulum is doubled, it becomes \( 2L \). Substitute \( 2L \) into the period formula:\[ T' = 2\pi \sqrt{\frac{2L}{g}} \]This is the new period of the pendulum after doubling the length.
3Step 3: Simplify the New Period Formula
Simplify the equation for the new period:\[ T' = 2\pi \sqrt{\frac{2L}{g}} = 2\pi \sqrt{2} \cdot \sqrt{\frac{L}{g}} \]Comparing this to the original period formula \( T = 2\pi \sqrt{\frac{L}{g}} \), we see that:\[ T' = \sqrt{2} \cdot T \]This shows that the new period is \( \sqrt{2} \) times the original period.
4Step 4: Calculate the Numerical Value
The square root of 2 is approximately 1.414. Therefore, doubling the length of the pendulum causes the period to increase by a factor of approximately 1.414.
Key Concepts
OscillationPendulum LengthAcceleration Due to Gravity
Oscillation
In the context of a pendulum, an oscillation refers to one complete movement of the pendulum bob from its starting point, to the opposite side, and back again to the starting point. It's a repeating, predictable motion.
This regular back-and-forth movement is what defines the stability and usefulness of pendulums in timekeeping, such as in grandfather clocks. The time it takes to complete one oscillation is known as the period.
This regular back-and-forth movement is what defines the stability and usefulness of pendulums in timekeeping, such as in grandfather clocks. The time it takes to complete one oscillation is known as the period.
- Each oscillation is consistent if initial conditions remain unchanged.
- Understanding oscillation is fundamental when exploring pendulum mechanics, including the effect of altering parameters like pendulum length.
Pendulum Length
Pendulum length plays a critical role in determining the period of the pendulum's oscillation. When referring to pendulum length, it means the distance from the pivot point to the center of mass of the pendulum bob.
According to the formula \[ T = 2\pi \sqrt{\frac{L}{g}} \] the period of the pendulum is directly related to the square root of its length. This means:
According to the formula \[ T = 2\pi \sqrt{\frac{L}{g}} \] the period of the pendulum is directly related to the square root of its length. This means:
- If the length increases, the period also increases.
- The relationship isn't linear; instead, it's a square root relationship, meaning doubling the length doesn't double the period—rather, it multiplies the period by \( \sqrt{2} \).
Acceleration Due to Gravity
The term "acceleration due to gravity" (symbolized as \( g \)) manifests as a crucial aspect of a pendulum's movement. This value, approximately 9.81 m/s² on Earth, affects how swiftly a pendulum accelerates towards the ground during its swing.
In the pendulum period formula, \[ T = 2\pi \sqrt{\frac{L}{g}} \], \( g \) is in the denominator, indicating:
In the pendulum period formula, \[ T = 2\pi \sqrt{\frac{L}{g}} \], \( g \) is in the denominator, indicating:
- A greater gravitational force decreases the pendulum's period, making oscillations quicker.
- Conversely, in environments with lesser gravity, a pendulum would oscillate slower.
Other exercises in this chapter
Problem 7
What is the wavelength of longitudinal waves in a coil spring with frequency \(7.50 \mathrm{~Hz}\) and velocity \(6.10 \mathrm{~m} / \mathrm{s}\) ?
View solution Problem 8
A wave generator produces 20 pulses in \(3.50 \mathrm{~s}\). (a) What is its period? (b) What is its frequency?
View solution Problem 9
A train traveling at a speed of \(4 \overline{0} \mathrm{~m} / \mathrm{s}\) approaches an observer at a station and sounds a \(55 \overline{0}\) -Hz whistle. Wh
View solution Problem 9
Find the frequency of a wave produced by a generator that emits 30 pulses in \(2.50 \mathrm{~s} .\)
View solution