The equation of motion for a spring-mass system excited by a harmonic force is
$M\ddot x + kx = F\cos \left( {\omega t} \right),$
Where M is the mass, K is the spring stiffness, F is the force amplitude and ω is the angular frequency of excitation. Resonance occurs when ω is equal to

Question

The equation of motion for a spring-mass system excited by a harmonic force is

$M\ddot x + kx = F\cos \left( {\omega t} \right),$

Where M is the mass, K is the spring stiffness, F is the force amplitude and ω is the angular frequency of excitation. Resonance occurs when ω is equal to

Prepp Team · Accepted Answer

Resonance in Spring-Mass Systems

The question asks about the condition for resonance in a spring-mass system when it is excited by a harmonic force. The given equation of motion for this system is:

\[M\ddot x + kx = F\cos \left( {\omega t} \right)\]

In this equation:

$M$ represents the mass of the system.
$k$ represents the spring stiffness.
$F$ represents the force amplitude of the external harmonic force.
$\omega$ represents the angular frequency of excitation of the external force.

Resonance Definition

Resonance is a critical phenomenon in vibration where the amplitude of oscillations in a system becomes maximum. This occurs when the frequency of the external exciting force matches the system's own inherent natural frequency. When the excitation frequency equals the natural frequency, the system absorbs maximum energy from the external source, leading to large amplitude vibrations.

Natural Frequency Derivation

To determine the condition for resonance, we first need to find the natural angular frequency of the spring-mass system. The natural frequency is the frequency at which the system would oscillate if it were disturbed and then left to vibrate freely, without any external forces or damping.

We consider the part of the equation of motion that describes free vibration, meaning we ignore the external harmonic force. So, the equation becomes:

\[M\ddot x + kx = 0\]

To simplify this, we can divide the entire equation by the mass $M$:

\[\ddot x + \frac{k}{M}x = 0\]

This equation is in the standard form for simple harmonic motion, which is generally written as $\ddot x + \omega_n^2 x = 0$. Here, $\omega_n$ represents the natural angular frequency of the system.

By comparing our derived equation $\ddot x + \frac{k}{M}x = 0$ with the standard form $\ddot x + \omega_n^2 x = 0$, we can directly identify the term for $\omega_n^2$:

\[\omega_n^2 = \frac{k}{M}\]

To find the natural angular frequency $\omega_n$, we take the square root of both sides:

\[\omega_n = \sqrt{\frac{k}{M}}\]

Resonance Condition

As discussed, resonance occurs when the angular frequency of excitation ($\omega$) is exactly equal to the natural angular frequency ($\omega_n$) of the system.

Therefore, the condition for resonance is:

\[\omega = \omega_n\]

Substituting the expression we found for $\omega_n$:

\[\omega = \sqrt{\frac{k}{M}}\]

Option Analysis

Let's check this result against the provided options for the value of $\omega$ at which resonance occurs:

Option 1: $\sqrt {\frac{M}{k}}$ - This is the inverse of the correct natural angular frequency.
Option 2: $\frac{1}{{2\pi }}\sqrt {\frac{k}{M}}$ - This represents the natural frequency in Hertz (cycles per second), not the angular frequency in radians per second.
Option 3: $2\pi \sqrt {\frac{k}{M}}$ - This is incorrect and does not represent the natural angular frequency.
Option 4: $\sqrt {\frac{k}{M}}$ - This matches our derived condition for resonance, which is when the excitation angular frequency $\omega$ equals the natural angular frequency $\omega_n$.

Therefore, resonance in the spring-mass system occurs when the angular frequency of excitation $\omega$ is equal to $\sqrt{\frac{k}{M}}$.