Question

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of the motion is 0.090 m, it takes the block 2.70 s to travel from \(x =\) 0.090 m to \(x = -\)0.090 m. If the amplitude is doubled, to 0.180 m, how long does it take the block to travel (a) from \(x =\) 0.180 m to \(x = -\)0.180 m and (b) from \(x =\) 0.090 m to \(x = -\)0.090 m?

Short answer

(a) 2.70 s; (b) 2.70 s.

Step-by-step solution

Determine the Frequency of Oscillation

For simple harmonic motion (SHM), the time taken for one complete cycle is the period, denoted by \( T \). When the amplitude is 0.090 m, it takes 2.70 s to go from \( x = 0.090 \text{ m} \) to \( x = -0.090 \text{ m} \). This is half the period \( T/2 \), so the full period is: \[ T = 2 \times 2.70 \text{ s} = 5.40 \text{ s} \]. The frequency \( f \) is the reciprocal of the period: \[ f = \frac{1}{T} = \frac{1}{5.40 \text{ s}} \].

Time to Travel Full Amplitude with Double Amplitude

When the amplitude doubles to 0.180 m, it doesn't affect the period. Since the period of one complete oscillation remains the same at 5.40 s, the time to travel from \( x = 0.180 \text{ m} \) to \( x = -0.180 \text{ m} \) is still half of the period: \( T/2 \). Thus, the time for this travel is: \[ \frac{5.40}{2} = 2.70 \text{ s} \].

Time to Travel Partial Amplitude with Double Amplitude

When the amplitude is doubled to 0.180 m, the time to travel between any two points within this amplitude (like from \( x = 0.090 \text{ m} \) to \( x = -0.090 \text{ m} \)) remains unchanged. This is because the trajectory between these specific points is still a quarter of a full oscillation. Hence, the time taken is the same as when the amplitude was 0.090 m, which is: \[ 2.70 \text{ s} \].

Key concepts

Understanding Frequency of Oscillation

Frequency of oscillation refers to how many complete cycles of motion an oscillating object undergoes per unit of time. It's a fundamental concept in simple harmonic motion (SHM). In SHM, the frequency is inversely related to the period of oscillation through the formula \( f = \frac{1}{T} \), where \( f \) is the frequency and \( T \) is the period.

This means that if you know the period, you can easily determine the frequency by taking the reciprocal of the period. For instance, when the amplitude of the oscillation was 0.090 m and took 2.70 s to cover from one extreme to the other, this represents half a cycle. The full period \( T \) thus becomes \( 5.40 \text{ s} \), making the frequency \( f = \frac{1}{5.40 \text{ s}} \approx 0.185 \text{ Hz} \).

The frequency remains constant regardless of changes in amplitude, because it is a property of the system's stiffness and mass, not the distance traveled during oscillations.

• Independent of amplitude.
• Determined by stiffness and mass of the system.
• Measured in Hertz (Hz).

Exploring the Period of Oscillation

The period of oscillation is the time it takes to complete one full cycle of motion in simple harmonic motion (SHM). In our scenario, the full cycle is captured when the block moves from its maximum positive amplitude to its maximum negative amplitude and back.

When the amplitude was initially 0.090 m, it took 2.70 s to move from one extreme to the other. This equates to half the oscillation period, meaning the full period is \( T = 5.40 \text{ s} \).

Interestingly, even with a double amplitude of 0.180 m, the period remains unchanged because period is independent of amplitude. The motion will still take 5.40 s to complete a full oscillation cycle.

• Calculated by doubling the half-cycle time.
• Remains constant despite changes in amplitude.
• Calculates timing for one full motion cycle.

Amplitude in SHM

Amplitude in simple harmonic motion (SHM) is the maximum extent of displacement from the equilibrium position. It is a measure of how far the block moves from its central rest position.

In the exercise, the amplitude initially was 0.090 m, and an increase to 0.180 m does not affect the period or frequency, but it does affect the maximum potential distance the block will travel in its oscillation.

Amplitude impacts only the trajectory path of the oscillating object. When it doubles, it signifies an increase in the maximum points of motion, but it does not change how fast the oscillations occur. Therefore, times initially calculated for travel between certain points within the amplitude range stay unchanged.

• Defines maximum displacement.
• Affects trajectory height, not the frequency or period.
• Change in amplitude affects energy potential.