5. What is the basis of vibrational spectroscopy. Derive the expression for natural frequency for vibration of a spring with spring constant K and mass m.
5. What is the basis of vibrational spectroscopy. Derive the expression for natural frequency for vibration of a spring with spring constant K and mass m.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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![**Question 5:**
*What is the basis of vibrational spectroscopy? Derive the expression for natural frequency for vibration of a spring with spring constant \(K\) and mass \(m\).*
### Explanation:
Vibrational spectroscopy involves studying the vibrational states of molecules. The molecules absorb energy at specific frequencies, leading to molecular vibrations observable through techniques like infrared and Raman spectroscopy.
### Derivation:
To find the natural frequency (\(\omega\)) of a spring-mass system:
- The system follows Hooke’s law and Newton’s second law.
- The force on the mass is given by \( F = -Kx \), where \( x \) is the displacement.
- From Newton’s second law, \( F = ma \), where \( a = \frac{d^2x}{dt^2} \).
Equating the two expressions for force:
\[ m \frac{d^2x}{dt^2} = -Kx \]
This differential equation can be rearranged as:
\[ \frac{d^2x}{dt^2} + \frac{K}{m}x = 0 \]
The solution to this equation is harmonic motion:
\[ x(t) = A \cos(\omega t + \phi) \]
Where:
- \( A \) is the amplitude,
- \( \phi \) is the phase,
- \( \omega \) is the angular frequency and is given by:
\[ \omega = \sqrt{\frac{K}{m}} \]
Thus, the natural frequency (\(f\)) is:
\[ f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{K}{m}} \]
This formula expresses the natural frequency of vibration for a spring system with stiffness \(K\) and mass \(m\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffae9e33c-257a-439a-bc57-ae888b0eb358%2Fc2a94af2-da5b-482e-aaf6-f0ba8460647c%2Fca6bn8k_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 5:**
*What is the basis of vibrational spectroscopy? Derive the expression for natural frequency for vibration of a spring with spring constant \(K\) and mass \(m\).*
### Explanation:
Vibrational spectroscopy involves studying the vibrational states of molecules. The molecules absorb energy at specific frequencies, leading to molecular vibrations observable through techniques like infrared and Raman spectroscopy.
### Derivation:
To find the natural frequency (\(\omega\)) of a spring-mass system:
- The system follows Hooke’s law and Newton’s second law.
- The force on the mass is given by \( F = -Kx \), where \( x \) is the displacement.
- From Newton’s second law, \( F = ma \), where \( a = \frac{d^2x}{dt^2} \).
Equating the two expressions for force:
\[ m \frac{d^2x}{dt^2} = -Kx \]
This differential equation can be rearranged as:
\[ \frac{d^2x}{dt^2} + \frac{K}{m}x = 0 \]
The solution to this equation is harmonic motion:
\[ x(t) = A \cos(\omega t + \phi) \]
Where:
- \( A \) is the amplitude,
- \( \phi \) is the phase,
- \( \omega \) is the angular frequency and is given by:
\[ \omega = \sqrt{\frac{K}{m}} \]
Thus, the natural frequency (\(f\)) is:
\[ f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{K}{m}} \]
This formula expresses the natural frequency of vibration for a spring system with stiffness \(K\) and mass \(m\).
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