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Chapter 14: Problem 10

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the block is at \(x=0.280 \mathrm{m},\) the acceleration of the block is \(-5.30 \mathrm{m} / \mathrm{s}^{2} .\) What is the frequency of the motion?

Short Answer

Expert verified
The frequency is approximately 0.692 Hz.

Step by step solution

01

Understand the Equation of Motion

In simple harmonic motion (SHM), the acceleration of an object is given by \[ a = -\omega^2 x \]where \(a\) is acceleration, \(\omega\) is the angular frequency, and \(x\) is the displacement from the equilibrium position. We're given that the acceleration \(a = -5.30 \ \text{m/s}^2\) when the displacement \(x = 0.280 \ \text{m}\).
02

Solve for Angular Frequency \(\omega\)

Take the equation for acceleration:\[ a = -\omega^2 x \]Plug in the values for \(a\) and \(x\): \[ -5.30 = -\omega^2 \times 0.280 \]Remove the negative signs and solve for \(\omega\^2\):\[ \omega^2 = \frac{5.30}{0.280} \approx \frac{53}{28} \approx 18.93 \]Calculate \(\omega\):\[ \omega = \sqrt{18.93} \approx 4.35 \ \text{rad/s} \]
03

Convert Angular Frequency to Frequency

The frequency \(f\) of the motion is related to the angular frequency \(\omega\) by the equation:\[ f = \frac{\omega}{2\pi} \]Substitute \(\omega = 4.35\) rad/s to find \(f\): \[ f = \frac{4.35}{2\pi} \approx \frac{4.35}{6.28} \approx 0.692 \ \text{Hz} \]
04

Final Verification

Check the calculations for any arithmetic errors and ensure that all units are consistent. After verification, ensure that the derived frequency \( f \approx 0.692 \ \text{Hz} \) is consistent with the given conditions and laws of SHM.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Angular Frequency
Angular frequency in simple harmonic motion (SHM) is a measure of how quickly an object moves through its cycle. It's denoted by the symbol \( \omega \) and is expressed in radians per second. This concept is crucial as it connects the displacement of a particle in SHM to its acceleration. In this scenario, angular frequency relates how a small block attached to an ideal spring oscillates back and forth.

To find the angular frequency, we use the formula for acceleration in SHM, which is given by \( a = - \omega^2 x \). Here, \(a\) represents acceleration, \(\omega\) is the angular frequency, and \(x\) is the displacement from equilibrium. By substituting the given values of acceleration and displacement into this formula, we can solve for \( \omega \).
  • Given: \( a = -5.30 \, \text{m/s}^2 \), \( x = 0.280 \, \text{m} \)
  • Equation: \( a = -\omega^2 x \)
  • Calculated \( \omega^2 = 18.93 \)
  • Angular frequency \( \omega \approx 4.35 \, \text{rad/s} \)
Acceleration in SHM
The acceleration in simple harmonic motion (SHM) is directly proportional to displacement but opposite in direction. This relation forms the core principle of SHM, indicating that as a particle moves further from its equilibrium position, it experiences a stronger pull back towards equilibrium.

Mathematically, this is captured by the equation \( a = - \omega^2 x \). Here:
  • \(a\) is the instantaneous acceleration.
  • \(x\) is the displacement from equilibrium.
  • The negative sign illustrates that acceleration acts in the opposite direction to displacement.
For instance, in this exercise, when the block is at 0.280 meters from equilibrium, its acceleration is \(-5.30 \mathrm{m/s}^2\). This signaling interplay between the distance of the block and its speed back to equilibrium vividly demonstrates the constant nature of pull in SHM.
Frequency Calculation
In physics, frequency indicates how often a repeating event occurs per unit time. In the context of SHM, it refers to how often an object like our block completes an oscillation cycle.

Frequency, \( f \), is calculated from angular frequency, \( \omega \), with the relationship \( f = \frac{\omega}{2\pi} \). Here's how it's linked:
  • \(\omega\) is in radians per second, while \(f\) is in hertz (Hz), or cycles per second.
  • Using the earlier determined \(\omega\approx 4.35 \text{ rad/s}\), we find the frequency:
  • \(f = \frac{4.35}{2\pi} \approx 0.692 \text{ Hz}\)
This calculation shows that the block oscillates back and forth approximately 0.692 times each second. A strong understanding of this conversion is vital for grasping how frequently such movements occur in everyday phenomena involving SHM.

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Most popular questions from this chapter

The point of the needle of a sewing machine moves in SHM along the \(x\) -axis with a frequency of 2.5 Hz. At \(t=0\) its position and velocity components are \(+1.1 \mathrm{cm}\) and \(-15 \mathrm{cm} / \mathrm{s},\) respectively. (a) Find the acceleration component of the needle at \(t=0\) . (b) Write equations giving the position, velocity, and acceleration components of the point as a function of time.

CALC A slender, uni- form, metal rod with mass \(M\) is pivoted without friction about an axis through its midpoint and perpendicular to the rod. A horizontal spring with force constant \(k\) is attached to the lower end of the rod, with the other end of the spring attached to a rigid support. If the rod is displaced by a small angle \(\theta\) from the vertical (Fig. \(P 14.97\) ) and released, show that it moves in angular SHM and calculate the period. (Hint: Assume that the angle \(\theta\) is small enough for the approximations \(\sin \theta \approx \theta\) and \(\cos \theta \approx 1\) to be valid. The motion is simple harmonic if \(d^{2} \theta / d t^{2}=-\omega^{2} \theta,\) and the period is then \(T=2 \pi / \omega .\) )

\(\mathrm{A} 0.500\) -kg mass on a spring has velocity as a function of time given by \(v_{x}(t)=-(3.60 \mathrm{cm} / \mathrm{s}) \sin \left[\left(4.71 \mathrm{s}^{-1}\right) t-\pi / 2\right]\) What are (a) the period; (b) the amplitude; (c) the maximum acceleration of the mass; (d) the force constant of the spring?

BIO Weighing a Virus. In February \(2004,\) scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the kind used in smallpox vaccine). The procedure involved measuring the frequency of oscillation of a tiny sliver of silicon (just 30 \(\mathrm{nm}\) long with a laser, first without the virus and then after the virus had attached itself to the silicon. The difference in mass caused a change in the frequency. We can model such a process as a mass on a spring. (a) Show that the ratio of the frequency with the virus attached \(\left(f_{\mathrm{S}+\mathrm{v}}\right)\) to the frequency without the virus \(\left(f_{\mathrm{S}}\right)\) is given by the formula \(\frac{f_{\mathrm{S}+\mathrm{V}}}{f_{\mathrm{S}}}=\frac{1}{\sqrt{1+\left(m_{\mathrm{V}} / m_{\mathrm{S}}\right)}}\) where \(m_{\mathrm{V}}\) is the mass of the virus and \(m_{\mathrm{S}}\) is the mass of the silicon sliver. Notice that is not necessary to know or measure the force constant of the spring. (b) In some data, the silicon sliver has a mass of \(2.10 \times 10^{-16} \mathrm{g}\) and a frequency of \(2.00 \times 10^{15} \mathrm{Hz}\) without the virus and \(2.87 \times 10^{14} \mathrm{Hz}\) with the virus. What is the mass of the virus, in grams and in femtograms?

Don't Miss the Boat. While on a visit to Minnesota ("Land of \(10,000\) Lakes"), you sign up to take an excursion around one of the larger lakes. When you go to the dock where the \(1500-\mathrm{kg}\) boat is tied, you find that the boat is bobbing up and down in the waves, executing simple harmonic motion with amplitude 20 \(\mathrm{cm} .\) The boat takes 3.5 \(\mathrm{s}\) s to make one complete up-and-down cycle. When the boat is at its highest point, its deck is at the same height as the stationary dock. As you watch the boat bob up and down, you (mass 60 \(\mathrm{kg}\) ) begin to feel a bit woozy, due in part to the previous night's dinner of lutefisk. As a result, you refuse to board the boat unless the level of the boat's deck is within 10 \(\mathrm{cm}\) of the dock level. How much time do you have to board the boat comfortably during each cycle of up-and-down motion?
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