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Chapter 14: Problem 8

When a 0.750 -kg mass oscillates on an ideal spring, the frequency is 1.33 Hz. What will the frequency be if 0.220 kg are (a) added to the original mass and (b) subtracted from the original mass? Try to solve this problem without finding the force constant of the spring.

Short Answer

Expert verified
The new frequency is 1.10 Hz with added mass and 1.71 Hz with subtracted mass.

Step by step solution

01

Understanding the Relation Between Mass and Frequency

The frequency of a mass-spring system is related to the mass and the spring constant by the formula \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \). Since we are not given the spring constant \( k \), and it is the same spring, the ratio of the frequencies for different masses can be used: \( \frac{f_1}{f_2} = \sqrt{\frac{m_2}{m_1}} \).
02

Calculating the New Frequency With Added Mass

For part (a), add 0.220 kg to the original mass, making the new mass \( m_2 = 0.750 + 0.220 = 0.970 \) kg. Using the ratio formula \( \frac{f_1}{f_2} = \sqrt{\frac{m_2}{m_1}} \) and solving for \( f_2 \), you get \( f_2 = \frac{f_1}{\sqrt{\frac{m_2}{m_1}}} = \frac{1.33}{\sqrt{\frac{0.970}{0.750}}} \). Calculate this to find \( f_2 \).
03

Computing Frequency With Decreased Mass

For part (b), subtract 0.220 kg from the original mass, so the new mass \( m_2 = 0.750 - 0.220 = 0.530 \) kg. Again, use the relation \( f_2 = \frac{f_1}{\sqrt{\frac{m_2}{m_1}}} = \frac{1.33}{\sqrt{\frac{0.530}{0.750}}} \). Calculate this to obtain \( f_2 \).
04

Final Calculation and Verification

For both cases (a) and (b), plug the calculated values into the respective formulas and simplify the square root to determine the new frequencies. Verify the calculations by ensuring the logical consistency that increased mass should decrease frequency, and decreased mass should increase frequency.

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Key Concepts

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

Mass-spring system
In physics, a mass-spring system is a simple model used to describe oscillatory motions, where a mass is attached to a spring. The mass moves back and forth under the influence of the spring's restoring force and its own inertia. This system is a classic example of simple harmonic motion, which is a type of periodic motion where the restoring force is directly proportional to the displacement.
  • The spring constant, often denoted as \( k \), is a measure of the spring's stiffness.
  • When the mass is displaced from its equilibrium position, the spring exerts a force that tries to bring the mass back to equilibrium.
  • This back-and-forth motion continues indefinitely in an ideal mass-spring system with no external forces like friction or air resistance.
Understanding how this system works helps in grasping more complex systems in mechanics. The interaction between the mass and the spring's properties determines how the system behaves during its oscillations.
Frequency calculation
The frequency of oscillation in a mass-spring system refers to how often the mass completes one full cycle of motion per second, and it is measured in Hertz (Hz). The fundamental frequency formula for a mass-spring system is given by:\[f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\]Where:
  • \( f \) is the frequency in Hz,
  • \( k \) is the spring constant in Newtons per meter,
  • \( m \) is the mass in kilograms.
This formula shows that the frequency depends on both the spring constant and the mass. To find the frequency when the mass changes without needing to re-calculate the spring constant, the ratio of frequencies for different masses can be used. This allows us to solve problems efficiently by comparing the change in mass and directly calculating the change in frequency.
  • As the mass increases, the frequency decreases because the mass takes longer to oscillate.
  • Conversely, when the mass decreases, the frequency increases.
This understanding is key in predicting how a mass-spring system reacts to changes in mass.
Relation between mass and frequency
The relation between mass and frequency in a mass-spring system can be easily understood through the frequency formula. As illustrated before, the frequency \( f \) is inversely proportional to the square root of the mass \( m \). This means that any increase in mass will result in a decrease in frequency, and similarly, any decrease in mass will cause the frequency to increase.
  • This inverse relationship can be expressed as: \( f \propto \frac{1}{\sqrt{m}} \).
  • If you add mass, you effectively slow down the system's ability to oscillate quickly.
  • On the other hand, subtracting mass makes it easier for the system to complete cycles more rapidly.
In practice, this helps in understanding phenomena like why adding cargo to a spring-loaded vehicle suspension makes it oscillate more slowly. By manipulating the mass attached to a spring, one can precisely control the frequency of oscillation, which is essential in designing systems that rely on predictable oscillatory behavior.

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Most popular questions from this chapter

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 50.0 \(\mathrm{cm} .\) She finds that the pendulum makes 100 complete swings in 136 s. What is the value of \(g\) on this planet?

CALC A 2.00 -kg bucket containing 10.0 \(\mathrm{kg}\) of water is hanging from a vertical ideal spring of force constant 125 \(\mathrm{N} / \mathrm{m}\) and oscillating up and down with an amplitude of 3.00 \(\mathrm{cm} .\) Suddenly the bucket springs a leak in the bottom such that water drops out at a steady rate of 2.00 \(\mathrm{g} / \mathrm{s}\) . When the bucket is half full, find (a) the period of oscillation and (b) the rate at which the period is changing with respect to time. Is the period getting longer or shorter? (c) What is the shortest period this system can have?

A 0.0200 -kg bolt moves with SHM that has an amplitude of 0.240 \(\mathrm{m}\) and a period of 1.500 \(\mathrm{s}\) . The displacement of the bolt is \(+0.240 \mathrm{m}\) when \(t=0 .\) Compute (a) the displacement of the bolt when \(t=0.500 \mathrm{s} ;(\mathrm{b})\) the magnitude and direction of the force acting on the bolt when \(t=0.500 \mathrm{s} ;\) (c) the minimum time required for the bolt to move from its initial position to the point where \(x=-0.180 \mathrm{m} ;\) (d) the speed of the bolt when \(x=-0.180 \mathrm{m}\) .

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?

CP SHM in a Car Engine. The motion of the piston of an automobile engine is approximately simple harmonic. (a) If the stroke of an engine (twice the amplitude) is 0.100 \(\mathrm{m}\) and the engine runs at 4500 \(\mathrm{rev} / \mathrm{min}\) , compute the acceleration of the piston at the endpoint of its stroke. (b) If the piston has mass \(0.450 \mathrm{kg},\) what net force must be exerted on it at this point? (c) What are the speed and kinetic energy of the piston at the mid- point of its stroke? (d) What average power is required to accelerate the piston from rest to the speed found in part (c)? (e) If the engine runs at 7000 rev/min, what are the answers to parts (b), (c), and (d)?
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