Frequency of Damped Vibrations In the chapter sound, my book states that the Frequency of damped vibrations is less than the natural frequency but I could not understand this because in damped vibrations the amplitude decreases and not the frequency.
May someone please explain this statement to me. I have googled this out but nowhere found the answer to my question rather there were confusing terms like damping coefficients etc. . My book also does not answer the "why" of my question.
P.S. I am a tenth grader so please avoid complicated terminologies. 
 A: Just think of stretching a mass on a spring and then releasing it. If the mass has a damper also attached, a force proportional to the speed of the mass will oppose the motion of the mass as it snaps away from you, thus slowing it compared to the case where there is no damper. Thus, the damped mass gets to the other side of the oscillation later than the undamped spring. This means the damped frequency is lower than the undamped frequency. 
The mathematical relationship is $\omega_\textrm{damped}=\omega_\textrm{undamped}\cdot \sqrt{1-\left(\frac{c}{2\sqrt{km}}\right)^2}$ where $c$ is the damper constant, $k$ is the spring constant, $m$ is the mass, $\omega_\textrm{undamped}$ is the natural frequency, and $\omega_\textrm{damped}$ is the damped frequency.
A: The equation of motion of an undamped spring is simply:
$$ma=-kx$$
Which is usually rewritten as:
$$m\ddot{x}+\omega_0^2x=0$$
With solution:
$$x=A\cos \omega_0t$$
Where the natural frequency $\omega_0$ is:
$$\omega_0=\sqrt{\frac{k}{m}}$$
For the damped case the equation of motion is:
$$ma=-kx-cv$$
Which is usually rewritten as:
$$m\ddot{x}+2\zeta\omega_0\dot{x}+\omega_0^2x=0$$
Where $\zeta$ is the damping ratio:
$$\zeta=\frac{c}{2\sqrt{mk}}$$
In the underdamped area ($\zeta<1$), which is the region of interest to the OP:
$$\omega_1=\omega_0\sqrt{1-\zeta^2}$$
Where $\omega_1$ is the damped frequency.
Obviously $\omega_1>\omega_0$.
