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A small fly of mass \(0.25 \mathrm{~g}\) is caught in a spider's web. The web oscillates predominately with a frequency of \(4.0 \mathrm{~Hz}\). (a) What is the value of the effective spring stiffness constant \(k\) for the web? ( \(b\) ) At what frequency would you expect the web to oscillate if an insect of mass \(0.50 \mathrm{~g}\) were trapped?

Short Answer

Expert verified
(a) The spring constant \( k \) is approximately \( 0.079 \mathrm{~N/m} \). (b) The web would oscillate at approximately \( 2.83 \mathrm{~Hz} \) with a \( 0.50 \mathrm{~g} \) insect.

Step by step solution

01

Understand Harmonic Motion

The fly in the spider's web causes the web to behave like a mass-spring system. The frequency of oscillation is given by the formula \[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\]where \( f \) is the frequency, \( k \) is the spring constant, and \( m \) is the mass attached to the spring.
02

Calculate Spring Constant for Initial Mass

To find the spring constant \( k \), rearrange the formula \[k = (2\pi f)^2 \cdot m\]Substituting the given values with the fly's mass \( m = 0.25 \mathrm{~g} = 0.00025 \mathrm{~kg} \) and frequency \( f = 4.0 \mathrm{~Hz} \):\[k = (2\pi \times 4.0)^2 \times 0.00025\]Calculate \( k \):\[k = (8\pi)^2 \times 0.00025 \approx 0.079 \mathrm{~N/m}\]
03

Calculate New Frequency for Increased Mass

Now, calculate the frequency for a mass of \( 0.50 \mathrm{~g} = 0.0005 \mathrm{~kg} \) using the same formula rearranged for frequency:\[ f' = \frac{1}{2\pi} \sqrt{\frac{k}{m'}}\]Substitute \( k = 0.079 \mathrm{~N/m} \) and \( m' = 0.0005 \mathrm{~kg} \):\[f' = \frac{1}{2\pi} \sqrt{\frac{0.079}{0.0005}}\]Calculate \( f' \):\[f' = \frac{1}{2\pi} \times 12.566 \approx 2.83 \mathrm{~Hz}\]

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Key Concepts

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

Mass-Spring System
In the realm of physics, a mass-spring system is a classic model to understand harmonic motion. This system consists of a mass attached to a spring, capable of moving back and forth. In our exercise, the spider's web behaves like this system. The mass of the tiny fly causes the web to oscillate. These repetitive movements are close to what we call simple harmonic motion.
For a mass-spring system, the ability to return back to its original position after being disturbed depends on the mass and the stiffness of the spring. Understanding this simple yet fascinating interaction helps us solve problems involving oscillations like the movement of the spider’s web.
Spring Constant
The spring constant, denoted as \( k \), is a measure of a spring's stiffness or rigidity. It tells us how much force is needed to stretch or compress the spring by a certain distance. In our problem, the web is equivalent to a spring where the spring constant needed to be calculated.
The calculation of the spring constant involved rearranging the frequency formula. By doing so, we used the fly's mass and the given frequency to find the value of \( k \). This resulted in a spring constant of approximately \( 0.079 \, \mathrm{N/m} \), suggesting that the web is quite delicate.
Oscillation Frequency
Oscillation frequency refers to how many times the mass on a spring completes a full cycle of movement in one second. In our scenario, this was first calculated for the fly which was trapped in the web. Its frequency was noted to be \( 4.0 \, \mathrm{Hz} \). This means the web oscillated four times every second with the fly caught in it.
We also calculated how the frequency would change if the mass trapped in the web were doubled. By substituting the new mass into the frequency formula, a reduced frequency of \( 2.83 \, \mathrm{Hz} \) was found. This shows that as the mass increases, the frequency of oscillation tends to decrease, meaning the system moves slower with a heavier mass.
Effective Spring Stiffness
Effective spring stiffness refers to the overall ability of a spring to resist deformation. In the context of the spider’s web, it indicates how strongly the web can resist changes in shape when a fly lands on it. This stiffness is quantified through the spring constant \( k \).
Having calculated the spring constant in our exercise, we can see how an increase in mass affects the stiffness perceived by the oscillating system. Even though the physical properties of the web don't change, the effect of the new mass alters how it oscillates. Understanding effective stiffness helps when predicting how different masses will impact oscillatory systems similar to this one.

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Most popular questions from this chapter

(II) A physical pendulum consists of an 85 -cm-long, 240 -g-mass, uniform wooden rod hung from a nail near one end (Fig, 38 ). The motion is damped because of friction in the pivot; the damping force is approximately proportional to \(d \theta / d t .\) The rod is set in oscillation by displacing it \(15^{\circ}\) from its equilibrium position and releasing it. After 8.0 \(\mathrm{s}\) the amplitude of the oscillation has been reduced to \(5.5^{\circ} .\) If the angular displacement can be written as \(\theta=A e^{-\gamma t} \cos \omega^{\prime} t,\) find \((a) \gamma,(b)\) the approximate period of the motion, and \((c)\) how long it takes for the amplitude to be reduced to \(\frac{1}{2}\) of its original value.

A \(0.650-\mathrm{kg}\) mass oscillates according to the equation \(x=0.25 \sin (5.50 t)\) where \(x\) is in meters and \(t\) is in seconds. Determine (a) the amplitude, (b) the frequency, (c) the period, (d) the total energy, and ( \(e\) ) the kinetic energy and potential energy when \(x\) is 15 \(\mathrm{cm} .\)

An object of unknown mass \(m\) is hung from a vertical spring of unknown spring constant \(k,\) and the object is observed to be at rest when the spring has extended by \(14 \mathrm{~cm} .\) The object is then given a slight push and executes SHM. Determine the period \(T\) of this oscillation.

(II) Tall buildings are designed to sway in the wind. In a 100 -km/h wind, for example, the top of the 110 -story Sears Tower oscillates horizontally with an amplitude of 15 \(\mathrm{cm}\) . The building oscillates at its natural frequency, which has a period of 7.0 s. Assuming SHM, find the maximum horizontal velocity and acceleration experienced by a Sears employee as she sits working at her desk located on the top floor. Compare the maximum acceleration (as a percentage) with the acceleration due to gravity.

(II) A 1.60 -kg object oscillates from a vertically hanging light spring once every 0.55 s. (a) Write down the equation giving its position \(y(+\) upward) as a function of time \(t,\) assuming it started by being compressed 16 \(\mathrm{cm}\) from the equilibrium position (where \(y=0 )\) , and released. \((b)\) How long will it take to get to the equilibrium position for the first time? (c) What will be its maximum speed? (d) What will be its maximum acceleration, and where will it first be attained?

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