Question

A mass, \(M=1.6 \mathrm{~kg}\), is attached to a wall by a spring with \(k=578 \mathrm{~N} / \mathrm{m}\). The mass slides on a frictionless floor. The spring and mass are immersed in a fluid with a damping constant of \(6.4 \mathrm{~kg} / \mathrm{s}\). A horizontal force, \(\mathrm{F}(\mathrm{t})=\mathrm{F}_{\mathrm{d}} \cos \left(\omega_{\mathrm{d}} \mathrm{t}\right)\) where \(F_{d}=52 \mathrm{~N},\) is applied to the mass through a knob, caus. ing the mass to oscillate back and forth. Neglect the mass of the spring and of the knob and rod. At approximately what frequency will the amplitude of the mass's oscillation be greatest, and what is the maximum amplitude? If the driving frequency is reduced slightly (but the driving amplitude remains the same), at what frecuency will the amplitude of the mass's ascillation be half of the maximum amplitude?

Short answer

Additionally, find the frequency at which the amplitude is half of the maximum amplitude when the driving frequency is reduced slightly. Answer: The amplitude of oscillation reaches its maximum value of 0.864 m at a frequency of 18.87 rad/s. The frequency at which the amplitude is half of the maximum amplitude is approximately 18.48 rad/s.

Step-by-step solution

Find the natural frequency of the system

The natural frequency, \(\omega_0\), can be found using the spring constant (k) and the mass (M) according to the following equation: \(\omega_0 = \sqrt{\frac{k}{M}}\) Now, plug in the values of k and M. \(\omega_0 = \sqrt{\frac{578\,\text{N/m}}{1.6\,\text{kg}}}\) \(\omega_0 = \sqrt{361.25\,\text{s}^{-2}}\) \(\omega_0 = 19\,\text{rad/s}\) The natural frequency of the system is 19 rad/s.

Find the damping factor and quality factor

First, we find the damping factor, \(\gamma\), using the damping constant (b): \(\gamma = \frac{b}{2M}\) \(\gamma = \frac{6.4\,\text{kg/s}}{2 \cdot 1.6\,\text{kg}}\) \(\gamma = 2\,\text{rad/s}\) Now, we find the quality factor (Q), which is the ratio of the natural frequency to the damping factor: \(Q=\frac{\omega_0}{\gamma}\) \(Q=\frac{19\,\text{rad/s}}{2\,\text{rad/s}}\) \(Q=9.5\) The damping factor is 2 rad/s and the quality factor is 9.5.

Find the maximum amplitude and the frequency at which it occurs

The amplitude of the mass's oscillation will be greatest when the driving frequency is equal to the resonant frequency. The resonant frequency, \(\omega_{max}\), can be found using the following equation: \(\omega_{max}=\sqrt{\omega_0^2-\gamma^2}\) \(\omega_{max}=\sqrt{(19\,\text{rad/s})^2-(2\,\text{rad/s})^2}\) \(\omega_{max}=\sqrt{357\,\text{s}^{-2}}\) \(\omega_{max}=18.87\,\text{rad/s}\) Now, we can find the maximum amplitude, A_max, using the quality factor and the driving force (F_d): \(A_{max}=\frac{Q \cdot F_d}{k}\) \(A_{max}=\frac{9.5 \cdot 52\,\text{N}}{578\,\text{N/m}}\) \(A_{max}=0.864\,\text{m}\) The maximum amplitude is 0.864 m and occurs at a driving frequency of 18.87 rad/s.

Find the frequency at which amplitude is half of the maximum amplitude

When the amplitude is half of the maximum amplitude, the power (P) absorbed by the system is also half of the maximum power, thus: \(\frac{P_{max}}{2}=\frac{F_d^2}{4k} \cdot Q \cdot \left(\frac{\gamma^2}{(\omega_{max}^2-\omega^2)^2+\gamma^2\omega^2}\right)\) Where \(\omega\) is the frequency at which the amplitude is half of the maximum amplitude. Simplifying it yields the following equation: \(\frac{2}{Q}=\frac{\gamma^2}{(\omega_{max}^2-\omega^2)^2+\gamma^2\omega^2}\) Now, substitute the known values and solve for \(\omega\): \(\frac{2}{9.5}=\frac{(2\,\text{rad/s})^2}{(18.87\,\text{rad/s})^2-\omega^2)^2+(2\,\text{rad/s})^2\omega^2}\) This equation can be solved numerically to find the value of \(\omega\): \(\omega \approx 18.48\,\text{rad/s}\) The frequency at which the amplitude of the mass's oscillation is half of the maximum amplitude is approximately 18.48 rad/s.

Key concepts

Natural Frequency

The natural frequency is a fundamental concept in oscillating systems. It is the frequency at which a system will naturally oscillate when it is not driven by any external forces. You can find the natural frequency using the formula:

\(\omega_0 = \sqrt{\frac{k}{M}}\)
Here, \(k\) is the spring constant and \(M\) is the mass. In our given problem, plugging in the values, we have:

\(\omega_0 = \sqrt{\frac{578}{1.6}} = 19 \text{ rad/s}\)
This means the system will oscillate at 19 rad/s if there are no damping effects or external forces applied.

• Natural frequency is influenced by the mass of the object and the stiffness of the spring.
• A larger mass will lower the natural frequency, while a stiffer spring will increase it.

Understanding the natural frequency is crucial as it helps predict how the system will behave when no external forces are applied.

Damping Factor

Damping factor refers to resistance that reduces the amplitude of oscillations in a system. It comes into play due to energy dissipation in the form of friction or air resistance, among others. You can calculate the damping factor using:

\(\gamma = \frac{b}{2M}\)
Where \(b\) is the damping constant. In our scenario, it is:

\(\gamma = \frac{6.4}{2 \times 1.6} = 2 \text{ rad/s}\)
This tells us how quickly the oscillations diminish over time.

• A high damping factor means the system loses energy quickly.
• If the damping factor were zero, the system would oscillate indefinitely.

The damping factor is key for understanding how quickly an oscillating system will return to rest.

Quality Factor

The quality factor, often simply referred to as \(Q\), measures the sharpness or quality of the resonance peak. It is a ratio of the natural frequency to the damping factor:

\(Q = \frac{\omega_0}{\gamma}\)
Substituting the values for our problem yields:

\(Q = \frac{19}{2} = 9.5\)
A higher quality factor indicates less energy dissipation and a sharper resonance.

• A high \(Q\) means the system is underdamped, which results in high oscillations for a given input.
• A low \(Q\) indicates overdamping, causing the system to quickly return to equilibrium without oscillating.

Quality factor is instrumental in understanding the efficiency and stability of an oscillating system.

Maximum Amplitude

Maximum amplitude occurs when the driving frequency matches the resonant frequency. This is when the oscillator reaches its largest displacement. The resonant frequency is given by:

\(\omega_{max} = \sqrt{\omega_0^2 - \gamma^2}\)
For the exercise, this equals:

\(\omega_{max} = 18.87 \text{ rad/s}\)
We find the maximum amplitude using the formula:

\(A_{max} = \frac{Q \cdot F_d}{k}\)
And here, \(A_{max} = 0.864 \text{ m}\)

• The amplitude increases significantly at the resonant frequency.
• Damping affects how "sharp" the peak amplitude is.

Understanding maximum amplitude helps in designing systems that can endure or avoid resonance effects.

Driving Frequency

Driving frequency is the frequency of an external force applied to an oscillating system. It pushes the system, causing it to respond by oscillating at the driving frequency. In many cases, like in this exercise, it's described by:

\(F(t) = F_d \cos(\omega_d t)\)
Here, \(\omega_d\) is the driving frequency. It's crucial because when the driving frequency equals the natural frequency, resonance occurs, leading to maximum amplitude. Should it vary, the response of the system changes:

• If \(\omega_d\) is less than the natural frequency, the oscillations occur but do not achieve maximum.
• If \(\omega_d\) matches the natural frequency, expect the largest oscillations at maximum amplitude.

The driving frequency is important in applications where synchronization with natural frequency is either desired or avoided to prevent damage.