Question

An oscillator consists of a block attached to a spring \((k=\) \(456 \mathrm{~N} / \mathrm{m}) .\) At some time \(t\), the position (measured from the equilibrium location), velocity, and acceleration of the block are \(x=0.112 \mathrm{~m}, v_{x}=-13.6 \mathrm{~m} / \mathrm{s}, a_{x}=-123 \mathrm{~m} / \mathrm{s}^{2} .\) Calcu- late \((a)\) the frequency, \((b)\) the mass of the block, and \((c)\) the amplitude of oscillation.

Short answer

The frequency of the oscillator is approximately \(\frac{1}{2\pi}\sqrt{\frac{456}{123}}\) Hz, the mass of the block is \(\frac{456}{123}\) kg, and the amplitude of the oscillation is roughly \(\sqrt{(0.112)^2 + (\frac{-13.6}{2\pi f})^2}\) m.

Step-by-step solution

Find the Frequency

First, calculate the frequency of the system using the formula \(f = \frac{1}{2\pi}\sqrt{\frac{k}{a_x}}\). Substituting the given values \(k = 456 \mathrm{~N/m}\) and \(a_x = -123 \mathrm{~m/s}^2\).Note, you will take the absolute value of \(a_x\) as acceleration can never be negative.The frequency \(f\) is therefore \(\frac{1}{2\pi}\sqrt{\frac{456}{123}}\).

Calculate the Mass

Next, find the mass of the block using the formula \(m = \frac{k}{a_x}\). Substitute the given values to get the mass \(m = \frac{456}{123}\).

Determine the Amplitude

Finally, calculate the amplitude of the oscillation using the formula \(A = \sqrt{x^2 + (\frac{v_x}{2\pi f})^2}\). Substitute the calculated frequency and given values \(x = 0.112 \mathrm{~m}\) and \(v_x = -13.6\mathrm{~m/s}\) (take absolute value) to find the amplitude \(A = \sqrt{(0.112)^2 + (\frac{-13.6}{2\pi f})^2}\).

Key concepts

Frequency Calculation in Harmonic Oscillators

In a harmonic oscillator, the frequency signifies how rapidly an object oscillates back and forth. It is crucial for understanding the system's behavior. In this exercise, the frequency is calculated using the formula:

• \( f = \frac{1}{2\pi}\sqrt{\frac{k}{a_x}} \)

where \(k\) is the spring constant and \(a_x\) is the acceleration. Importantly, remember to take the absolute value of the acceleration, as acceleration in physics is considered a magnitude or absolute value. Inserting the values, we find:

• \( f = \frac{1}{2\pi}\sqrt{\frac{456}{123}} \)

This formula stem from the nature of simple harmonic motion and reflects how interconnected the system's parameters are to the oscillation frequency.

Mass Determination of the Oscillator Block

Determining the mass of an oscillator block is directly tied to understanding how a force influences the oscillator. The relation of the spring constant and acceleration gives us insight into mass using this formula:

• \( m = \frac{k}{a_x} \)

By plugging in the values \(k = 456 \text{ N/m}\) and \(a_x = 123 \text{ m/s}^2\), we can deduce that:

• \( m = \frac{456}{123} \text{ kg} \)

The mass results from how much force is applied by the spring to alter the motion of the block. It is vital as it directly impacts the motion’s frequency and amplitude.

Calculating the Amplitude of Oscillation

Amplitude in a harmonic oscillator is indicative of the maximum extent of the oscillation. A larger amplitude means the object moves further from the center point. The formula for amplitude in our context is:

• \( A = \sqrt{x^2 + \left(\frac{v_x}{2\pi f}\right)^2} \)

Here, \(x\) is the initial displacement, \(v_x\) is the velocity (in absolute terms), and \(f\) is the frequency we've previously calculated. Inserting the provided values:

• \( A = \sqrt{(0.112)^2 + \left(\frac{-13.6}{2\pi f}\right)^2} \)

This computation results in the distance the oscillator reaches from its equilibrium point during the motion. Understanding amplitude helps comprehend the energy in the system as well as potential energy maxima.