If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Bonds as springs

Learn how frequency of bond vibration can be compared to the oscillation of a spring by using Hooke's law. Explore how the strength of the bond and the mass of the atoms affect the frequency of vibration. The concept of reduced mass is introduced for situations where both atoms in a bond are moving. Created by Jay.

## Want to join the conversation?

• At , it is said that stronger bonds vibrate faster. So does a C double bond vibrate faster than a C-H bond?
• No. A C-H bond vibrates almost twice as fast as a C=C bond — 3000 cm⁻¹ vs. 1700 cm⁻¹.
If you have atoms of the same type, the stronger bond vibrates faster.
C≡C > C=C > C-C (2200 cm⁻¹ > 1700 cm⁻¹ > 1200 cm⁻¹)
The vibrational frequency of a vibrating two-body system depends on both the force constant and the reduced mass of the system: ν ∝ √(k/µ).
The force constant depends on the strength of the bond. A C=C bond is about 1.5 times as strong as a C=C bond. That would make it vibrate faster than a C-H bond.
But µ(C-C) = 6, while µ(C-H) = 0.92. The reduced mass of a C-C pair is about 6.5 times that of a C-H pair. The greater reduced mass makes the C=C bond vibrate much more slowly than a C-H bond.
A C-H bond vibrates faster by a factor of about √(6.5/1.5) ≈ 2.
• I don't understand why at we are we using a reduced mass instead of say, m1 + m2
• In a vibration, it's not just one atom that moves back and forth. The centre of mass must stay in the same place.
Thus, as one atom moves in one direction, the other atom must move in the opposite direction.
This is what the physicists call a "two-body vibration" problem, in which the two atoms vibrate about a common centre of mass.
The position of the centre of mass is determined by the reduced mass of the system.
• At , if increasing the reduced mass results in a lower frequency of vibration, why does an O-H bond stretch at a higher wavenumber than C-H? Shouldn't the increased reduced mass of O-H make the wavenumber go lower?
• The frequency also depends on the force constant k, which corresponds roughly to the bond strength.
An OH bond is stronger than a CH bond. Its effect overpowers the effect of the reduced mass.
• How true is the assumption that a double bond is twice the strength of a single bond?
• Would it be correct to assume that sound wave resonance works on the same principal? As a musician, main instrument being the guitar, I see it as being analogous to guitar string physics and for me personally, it would seem to be an easy way to make sense of it all. A heavier(thicker) string has a lower frequency and the tighter the string, the higher the pitch. If so, does harmonic resonance play a part in the resulting noise in the spectrograph?
• Can someone please direct me to the full calculations which resulted to the equation presented at (a name, or a link)? Thanks.
• I can understand your interest here but it probably wouldn't be very helpful to know how that equation was derived. This is because in this series of videos Jay is explaining everything in terms of classical physics but if you were studying this at a much deeper level then you would need to learn about the quantum physics of the vibrations and this would involve different, and more complex, equations which are derived from the Schrodinger equation. The classical physics approach serves as a good model for the quantum physics of vibrations but it has its limitations. You would not gain anything to help you interpret IR spectra by understanding how the classical equation at is derived.
• I don't quite understand what the K-value is representing in regards to the bonds between atoms. When talking about springs, k is the spring constant, right, meaning the stiffness of the spring and what not. What is it really for bonds though?
• I'm guessing that it is related to both the density and focus of the electron "cloud(s)" that form the bond(s).
(I'm thinking of the masses as representing the nuclei with the spring representing the electron "cloud".)
• How should I think of this "bond stretching" in terms of the MOT ? Would it be accurate to think of it as a certain "deformation" of the molecular orbital ( the electron density around the bonded atoms in the molecule) ?
And also how should I think of this in terms of the VBT where bonds are just overlapping between the orbitals ?