Question

If an object on a horizontal, frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If it is displaced 0.120 m from its equilibrium position and released with zero initial speed, then after 0.800 s its displacement is found to be 0.120 m on the opposite side, and it has passed the equilibrium position once during this interval. Find (a) the amplitude; (b) the period; (c) the frequency.

Short answer

Amplitude is 0.120 m, period is 1.600 s, and frequency is 0.625 Hz.

Step-by-step solution

Understanding Amplitude

The amplitude of oscillation is the maximum displacement from the equilibrium position. In this case, the object is initially displaced by 0.120 m from equilibrium and reaches the same displacement on the opposite side, indicating the amplitude is 0.120 m.

Identifying the Period

The period (T) is the time it takes to complete one full cycle of oscillation. Since the object moves from 0.120 m on one side to 0.120 m on the other side and passes equilibrium once, this is half a cycle. Thus, the time for a full cycle is twice 0.800 s: \[ T = 2 \times 0.800 \text{ s} = 1.600 \text{ s}. \]

Calculating the Frequency

The frequency (f) is the number of cycles per second and is the reciprocal of the period. Calculate frequency using the formula: \[ f = \frac{1}{T} = \frac{1}{1.600} \approx 0.625 \text{ Hz}. \]

Key concepts

Amplitude

In the context of Simple Harmonic Motion (SHM), the amplitude is a crucial concept for understanding the behavior of oscillating systems. The amplitude is defined as the maximum extent of displacement from the equilibrium position. In essence, it tells us how far the object moves from its central point in either direction during its motion.
In our exercise, we observe that the object reaches a displacement of 0.120 meters on one side of the equilibrium position, then moves to the exact same distance on the opposite side. This uniformity in displacement on both sides directly conveys that the amplitude is 0.120 meters. Amplitude is always a positive number and represents the peak value of the displacement vector in SHM cycles. Think of it as the 'height' of the oscillation, defining the limits within which the object will continue to oscillate back and forth.
An important point to remember is that amplitude is not influenced by the period or frequency of the oscillation. It purely depends on how far the object was initially displaced from its rest position.

Period of Oscillation

The period of oscillation (T) is another fundamental concept in understanding SHM. It refers to the time needed to complete one full cycle of motion. Imagine watching a pendulum swing; the period would be the time it takes to swing and return to the starting point.
In our specific scenario, we find that the object takes 0.800 seconds to move from the initial displacement of 0.120 meters to the opposite displacement of -0.120 meters and back through the equilibrium position. It crosses the starting line, completing half of a full oscillation cycle. Since this half-cycle takes 0.800 seconds, a full cycle (period) would take twice as long.
Thus, the period, T, is calculated as:

• T = 2 × 0.800 s = 1.600 s.

The period is vital because it defines how quickly or slowly the oscillation process occurs, yet it remains consistent across all cycles, provided the system is undisturbed. It is also inversely related to frequency, but more on that in the next section!

Frequency

Frequency (f) in Simple Harmonic Motion describes how many cycles of the oscillation occur in one second. This is where the unit 'Hertz' (Hz) comes into play, representing cycles per second. For students diving into physics, frequency is crucial because it provides insight into the rate of oscillation, linking back to the concept of time.
From our problem, we know the period of oscillation is 1.600 seconds. Frequency and period have an inverse relationship, expressed through the formula:

• f = \( \frac{1}{T} \)

Plugging in the period, the frequency is calculated as:

• f = \( \frac{1}{1.600} \approx 0.625 \text{ Hz} \)

This frequency tells us that approximately 0.625 cycles are completed each second. In essence, frequency shifts our perspective from asking how long one cycle takes (period) to asking how many cycles are completed in a given timeframe, thus providing a different but connected way of understanding oscillations.