Question

An oscillator consists of a block of mass \(512 \mathrm{~g}\) connected to a spring. When set into oscillation with amplitude \(34.7 \mathrm{~cm}\), it is observed to repeat its motion every \(0.484 \mathrm{~s}\). Find \((a)\) the period, \((b)\) the frquency, \((c)\) the angular frequency, \((d)\) the force constant, \((e)\) the maximum speed, and \((f)\) the maximum force exerted on the block.

Short answer

The period is 0.484 s, the frequency is 2.066 Hz, the angular frequency is 12.98 rad/s, the force constant is 86.64 N/m, the maximum speed is 4.50 m/s, and the maximum force is 30.02 N.

Step-by-step solution

Calculate The Period

The period (T) of the oscillation is simply the time it takes for one complete cycle of the motion. Given in the problem, this is \(T = 0.484 \mathrm{~s}\). No calculation is necessary here as the period is already given.

Calculate The Frequency

The frequency (f) of an oscillation is the reciprocal of the period. This can be calculated with the formula \(f = 1/T\). So \(f = 1/0.484 \approx 2.066 \mathrm{~Hz}\).

Calculate The Angular Frequency

The angular frequency (ω) is related to the frequency by \(ω = 2πf\), so, substituting the known value of f, \(ω = 2π × 2.066 \approx 12.98 \mathrm{rad/s}\).

Calculate The Force Constant

The force constant (k) of the spring in the oscillator can be calculated using the formula \(k = mω^2\), where m is the mass of the block. Make sure to convert the mass from grams to kilograms before calculating. \(k = 0.512 \mathrm{(kg)} × (12.98 \mathrm{rad/s})^2 \approx 86.64 \mathrm{N/m}\).

Calculate The Maximum Speed

The maximum speed (vmax) occurs when the potential energy is minimum, that is when the displacement (x) from the equilibrium position is zero. vmax can be calculated using the formula \(vmax = ω \cdot amplitude\). Therefore, \(vmax = 12.98 \mathrm{rad/s} × 0.347 \mathrm{m} \approx 4.50 \mathrm{m/s}\).

Calculate The Maximum Force

The maximum force (Fmax) is the force exerted when the block is at maximum displacement, i.e., at an extreme position. It is given by the formula \(Fmax = k \cdot amplitude\). Therefore, \(Fmax = 86.64 \mathrm{N/m} × 0.347 \mathrm{m} \approx 30.02 \mathrm{N}\).

Key concepts

Oscillator Physics

Simple harmonic motion (SHM) is fundamental to understanding oscillator physics. It describes the motion of objects, such as springs and pendulums, which return to an equilibrium position because of a restoring force. The force is directly proportional to the displacement from equilibrium, following Hooke's Law, which states \( F = -kx \), where \( F \) is the restoring force, \( k \) is the force constant or spring constant, and \( x \) is the displacement.

An oscillator in SHM has a characteristic motion of swinging back and forth or moving up and down with a specific, repeatable pattern, which can be graphically represented as a sine or a cosine wave. The mass attached to the spring in the given exercise undergoes this exact type of motion.

Spring Constant Calculation

The spring constant, denoted \( k \), measures the stiffness of a spring and is a factor in determining how a spring will stretch or compress under a given force. Calculating \( k \) requires knowing the system's mass \( m \) and the angular frequency \( \omega \), as illustrated by the formula \( k = m\omega^2 \). Remember to always convert mass into the correct units (kilograms in SI units) before using it in calculations. In our example, after converting the mass of the block to kilograms, we calculate \( k \) using the angular frequency obtained from the given frequency of the SHM.

Angular Frequency

Angular frequency \( \omega \) plays a pivotal role in describing the rate of oscillation in terms of angles in a time interval. It is expressed in radians per second (\(\mathrm{rad/s}\)) and is closely related to the linear frequency \( f \) via the equation \( \omega = 2\pi f \). This relationship implies that the angular frequency represents the number of radians the oscillating particle goes through per second. In the exercise solution, we see that angular frequency is determined using the frequency, which is the reciprocal of the period of the oscillator.

Oscillation Period

The period \( T \) of an oscillator in SHM is the time it takes to complete one full cycle of oscillation. It's connected to the frequency \( f \) of the oscillation, where \( T = 1/f \) and is measured in seconds. When examining an oscillator motion graph, the period is the horizontal length from one peak to the next or from any point in one cycle to the identical point in the next cycle. It's critical to point out that \( T \) is a constant for a given system and does not change unless the system's properties change.

Maximum Speed and Force

The maximum speed \( vmax \) of an object undergoing SHM occurs as it passes through the equilibrium point, where kinetic energy is at its peak. The formula \( vmax = \omega \cdot amplitude \) illustrates this relationship, reflecting that \( vmax \) depends on both the angular frequency and the amplitude of the motion. Maximum force \( Fmax \) occurs at the maximum displacement, which corresponds to the amplitude of the motion. It's determined with Hooke's Law, \( Fmax = k \cdot amplitude \) and represents the peak restoring force exerted by the spring on the mass. Understanding these maximum values helps in understanding the energy transfer within the SHM system.