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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?

Short Answer

Expert verified
The mass of the virus is approximately 2.06 x 10^-16 grams or 206 femtograms.

Step by step solution

01

Understand the Problem

We are given a scenario where the frequency of oscillation changes when the virus attaches to a silicon sliver. Our goal is to find the mass of the virus based on the change in frequency.
02

Identify the Given Formulas

We need to show that the frequency ratio formula \(\frac{f_{S+V}}{f_S} = \frac{1}{\sqrt{1 + \left(\frac{m_V}{m_S}\right)}}\). This requires understanding that frequency \(f\) for a mass-spring system is \(f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\), where \(k\) is the spring constant and \(m\) is the mass. However, we don't need \(k\) for this problem.
03

Derive the Frequency Ratio Formula

Starting from \(f_{S} = \frac{1}{2\pi} \sqrt{\frac{k}{m_S}}\) and \(f_{S+V} = \frac{1}{2\pi} \sqrt{\frac{k}{m_S + m_V}}\), we find the ratio:\[ \frac{f_{S+V}}{f_S} = \frac{\sqrt{\frac{k}{m_S + m_V}}}{\sqrt{\frac{k}{m_S}}} = \sqrt{\frac{m_S}{m_S + m_V}} \]Which simplifies to \(\frac{1}{\sqrt{1 + \left(\frac{m_V}{m_S}\right)}}\).
04

Plug in Given Values for mass of virus

Given:- \(f_S = 2.00 \times 10^{15}\, \text{Hz}\)- \(f_{S+V} = 2.87 \times 10^{14}\, \text{Hz}\)- \(m_S = 2.10 \times 10^{-16}\, \text{g}\)Find \(m_V\) using \(\frac{f_{S+V}}{f_S} = \frac{1}{\sqrt{1 + \left(\frac{m_V}{m_S}\right)}}\).
05

Calculate the Frequency Ratio

Calculate the frequency ratio:\[ \frac{f_{S+V}}{f_S} = \frac{2.87 \times 10^{14}}{2.00 \times 10^{15}} = 0.1435 \]
06

Solve for mass of virus \(m_V\)

Set up the equation:\[ 0.1435 = \frac{1}{\sqrt{1 + \frac{m_V}{2.10 \times 10^{-16}}}} \]Square both sides:\[ 0.02059 = \frac{1}{1 + \frac{m_V}{2.10 \times 10^{-16}}} \]Rearrange to find \(m_V\):\[ m_V = (1 - 0.02059) \times 2.10 \times 10^{-16} \]\[ m_V = 0.97941 \times 2.10 \times 10^{-16} \]\[ m_V \approx 2.06 \times 10^{-16} \text{ grams} \]
07

Convert to Femtograms

Convert the mass from grams to femtograms using the conversion \(1 \text{g} = 10^{15} \text{fg}\):\[ m_V \approx 2.06 \times 10^{-16} \times 10^{15} = 206 \text{ fg} \]

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Key Concepts

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

Mass-spring system
The mass-spring system is a fundamental concept in physics used to describe oscillatory motion and dynamics. In such a system, a mass is attached to a spring. When displaced from its equilibrium position and released, the mass experiences a restoring force from the spring. This force tries to bring it back to equilibrium, causing oscillations.

Key Characteristics of Mass-Spring Systems:
  • Restoring Force: Directly proportional to displacement, given by Hooke's law: \( F = -kx \) where \( k \) is the spring constant and \( x \) is the displacement.
  • Equilibrium position: The point where the net force on the mass is zero. If undisturbed, the mass will remain at this position.
  • Oscillatory Motion: The mass moves back and forth around the equilibrium position, which is typical of simple harmonic motion.
The frequency of oscillation in a mass-spring system depends on the spring constant and the mass. Understanding the mass-spring system underlies many natural and technological processes, from mechanical watches to molecular vibrations.
Oscillation frequency
In a mass-spring system, the oscillation frequency is a crucial property. It tells us how fast the system oscillates, measured in Hertz (Hz), or cycles per second.

For a simple mass-spring system, the frequency \( f \) is determined by the formula:\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]where:
  • \( k \): Spring constant, indicating stiffness of the spring.
  • \( m \): Mass attached to the spring.
Oscillation Frequency's Importance:
  • Stability and Dynamics: In systems like buildings or bridges, knowing the natural oscillation frequencies can help engineers ensure they're not resonating with external forces (like wind).
  • Physics and Engineering: Many devices and natural phenomena rely on resonant frequencies, from radio transmitters to musical instruments.
Understanding oscillation frequency helps predict system behavior under various conditions, just as used in the virus mass measurement problem.
Virus mass measurement
Measuring virus mass, such as the vaccinia virus, can be achieved using changes in oscillation frequency. The tiny mass change when a virus attaches to a finely tuned system, like a silicon sliver modeled as a mass-spring system, alters the frequency.

Steps in Measuring Virus Mass:
  • Initial Measurement: Determine the oscillation frequency of an isolated silicon sliver.
  • Attachment of Virus: Allow the virus to attach to the sliver, adding its mass.
  • Frequency Change: Measure the new, lower frequency due to the added mass.
Using the formula:\[ \frac{f_{\text{S+V}}}{f_{\text{S}}} = \frac{1}{\sqrt{1 + \left(\frac{m_{\text{V}}}{m_{\text{S}}}\right)}} \]where \( f_{\text{S+V}} \) and \( f_{\text{S}} \) are the frequencies with and without the virus, respectively, and \( m_{\text{V}} \) and \( m_{\text{S}} \) represent the virus's and sliver's masses, we can solve for the virus's mass. This allows precise measurements without directly handling subatomic scales, making it a powerful technique for virology.

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Most popular questions from this chapter

A 1.50 -kg mass on a spring has displacement as a function of time given by the equation $$x(t)=(7.40 \mathrm{cm}) \cos \left[\left(4.16 \mathrm{s}^{-1}\right) t-2.42\right]$$ Find (a) the time for one complete vibration; (b) the force constant of the spring; (c) the maximum speed of the mass; (d) the maximum force on the mass; (e) the position, speed, and acceleration of the mass at \(t=1.00 \mathrm{s} ;\) (f) the force on the mass at that time.

A proud deep-sea fisherman hangs a 65.0 -kg fish from an ideal spring having negligible mass. The fish stretches the spring 0.120 \(\mathrm{m}\) . (a) Find the force constant of the spring. The fish is now pulled down 5.00 \(\mathrm{cm}\) and released. (b) What is the period of oscillation of the fish? (c) What is the maximum speed it will reach?

An object is undergoing SHM with period 1.200 s and amplitude 0.600 \(\mathrm{m}\) . At \(t=0\) the object is at \(x=0\) and is moving in the negative \(x\) -direction. How far is the object from the equilibrium position when \(t=0.480 \mathrm{s} ?\)

BIO Weighing Astronauts. This procedure has actually been used to "weigh"" astronauts in space. A 42.5 -kg chair is attached to a spring and allowed to oscillate. When it is empty, the chair takes 1.30 s to make one complete vibration. But with an astronaut sitting in it, with her feet off the floor, the chair takes 2.54 s for one cycle. What is the mass of the astronaut?

The Silently Ringing Bell Problem. A large bell is hung from a wooden beam so it can swing back and forth with negligible friction. The center of mass of the bell is 0.60 m below the pivot, the bell has mass 34.0 \(\mathrm{kg}\) , and the moment of inertia of the bell about an axis at the pivot is 18.0 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) The clapper is a small, \(1.8-\mathrm{kg}\) mass attached to one end of a slender rod that has length \(L\) and negligible mass. The other end of the rod is attached to the inside of the bell so it can swing freely about the same axis as the bell. What should be the length \(L\) of the clapper rod for the bell to ring silently-that is, for the period of oscillation for the bell to equal that for the clapper?

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