/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 An oxygen atom at a particular s... [FREE SOLUTION] | 91Ó°ÊÓ

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An oxygen atom at a particular site within a DNA molecule can be made to execute simple harmonic motion when illuminated by infrared light. The oxygen atom is bound with a spring-like chemical bond to a phosphorus atom, which is rigidly attached to the DNA backbone. The oscillation of the oxygen atom occurs with frequency \(f =\) 3.7 \(\times\) 10\(^{13}\) Hz. If the oxygen atom at this site is chemically replaced with a sulfur atom, the spring constant of the bond is unchanged (sulfur is just below oxygen in the Periodic Table). Predict the frequency after the sulfur substitution.

Short Answer

Expert verified
The new frequency is approximately 2.62 × 10¹³ Hz.

Step by step solution

01

Understanding the System

We are dealing with a simple harmonic oscillator, where the oxygen atom, which has a certain mass, oscillates with a frequency determined by the spring constant of the bond. After replacing the oxygen with sulfur, the mass changes, but the spring constant remains the same.
02

Identifying Given Values

The initial frequency of oscillation for oxygen is given as \(f = 3.7 \times 10^{13}\) Hz. We also know that the spring constant \(k\) stays the same upon replacing oxygen with sulfur.
03

Relating Frequency to Mass and Spring Constant

For a simple harmonic oscillator, the frequency \(f\) is related to the spring constant \(k\) and the mass \(m\) by the equation: \(f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\). We need to use this relationship to find the new frequency when sulfur replaces oxygen.
04

Calculating Initial Mass Ratio

The initial frequency of the oxygen atom is \(f_O = \frac{1}{2\pi}\sqrt{\frac{k}{m_O}}\). We rearrange to find \(\sqrt{k} = 2\pi f_O \sqrt{m_O}\). This expression will help us calculate the new frequency for sulfur, assuming mass substitution.
05

Substituting New Mass of Sulfur

Let the mass of sulfur be \(m_S\). The new frequency can be found by \(f_S = \frac{1}{2\pi}\sqrt{\frac{k}{m_S}}\). So \(f_S = \frac{f_O}{\sqrt{m_S/m_O}}\), since the ratio \(\sqrt{k}\) remains the same due to unchanged spring constant.
06

Calculating the Frequency Ratio

We find the masses of oxygen \(m_O\) and sulfur \(m_S\) on the periodic table. Oxygen (\(O\)) has a relative atomic mass of approximately 16, and sulfur (\(S\)) has a relative atomic mass of approximately 32. Calculate the ratio \(\sqrt{m_O/m_S} = \sqrt{16/32} = \sqrt{0.5} = \sqrt{1/2} = 1/\sqrt{2}\).
07

Finding the New Frequency

The new frequency \(f_S\) will then be \(f_S = f_O \times \frac{1}{\sqrt{2}} = 3.7 \times 10^{13} \times \frac{1}{\sqrt{2}}\). This gives \(f_S \approx 2.62 \times 10^{13}\) Hz.

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Key Concepts

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

Frequency Calculation
In the realm of simple harmonic motion, frequency is a pivotal concept. Frequency, denoted by \( f \), is essentially how often an oscillation repeats in a second, measured in Hertz (Hz). For a simple harmonic oscillator, the frequency is calculated using the formula:
  • \( f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \)
The variables here include the spring constant \( k \) and the mass \( m \). Understanding this relationship helps predict behavior if changes, such as mass substitution, occur. To maintain the same oscillatory behavior, both mass and spring constant affect the frequency. This formula is derived from the motion characteristics of the system, showing how frequency ultimately depends on these two factors. Substituting different values for mass or spring constant allows us to calculate new frequencies, which is crucial in understanding how these oscillators work in nature.
Mass Substitution
When we replace an atom in a system, like substituting oxygen with sulfur, the system's mass changes. This substitution directly influences the frequency of oscillation in a simple harmonic motion without affecting the spring constant. The initial mass of oxygen, denoted \( m_O \), differs from that of sulfur, \( m_S \). Mass substitution requires recalculating oscillation frequency using altered mass values while keeping the spring constant the same.
  • Oxygen's relative atomic mass is approximately 16.
  • Sulfur's relative atomic mass is approximately 32.
The relationship between mass and frequency can be adjusted as \( f_S = f_O \times \frac{1}{\sqrt{m_S/m_O}} \). This equation reflects the inverse square root relationship between mass ratio and frequency, meaning a heavier mass results in a lower frequency of oscillation. After calculating the mass ratio, you can predict the new frequency of the system.
Spring Constant
The spring constant, represented as \( k \), is crucial in determining the frequency in simple harmonic motion. It describes the strength of the spring bond, indicating how quickly or slowly an atom in a molecule will return to its equilibrium position after being displaced.
  • The spring constant remains unchanged during mass substitution.
  • A higher spring constant typically results in a higher frequency, as it indicates a stiffer spring bond that resists deformation more effectively.
Since the spring constant was unchanged while substituting oxygen with sulfur in this problem, the calculation of frequency changes relies solely on mass differences. Understanding the role of the spring constant helps explain why the motion's nature, as in this case, primarily relies on any changes in mass. The predictability of \( k \) emphasizes its role acting independent of mass in determining frequency outcomes when only the mass in the system varies.

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Most popular questions from this chapter

A rectangular block of wood floats in a calm lake. Show that, if friction is ignored, when the block is pushed gently down into the water and then released, it will then oscillate with SHM. Also, determine an equation for the force constant.

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An energy-absorbing car bumper has a spring constant of 410 kN\(/\)m. Find the maximum compression of the bumper if the car, with mass 1300 kg, collides with a wall at a speed of 2.0 m\(/\)s (approximately 5 mi\(/\)h ).

A fisherman's scale stretches 3.6 cm when a 2.4-kg fish hangs from it. (\(a\)) What is the spring stiffness constant and (\(b\)) what will be the amplitude and frequency of oscillation if the fish is pulled down 2.1 cm more and released so that it oscillates up and down?

Two strings on a musical instrument are tuned to play at 392 Hz (G) and 494 Hz (B). (\(a\)) What are the frequencies of the first two overtones for each string? (\(b\)) If the two strings have the same length and are under the same tension, what must be the ratio of their masses (\(m_G/m_B\))? (\(c\)) If the strings, instead, have the same mass per unit length and are under the same tension, what is the ratio of their lengths (\(\ell_G/\ell_B\))? (\(d\)) If their masses and lengths are the same, what must be the ratio of the tensions in the two strings?

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