a. Can you write the basic reason of resonance? Is there any role of natural frequency in thisphenomenon or not?b. Generally it is accepted about the slope at any point on the displacement graph of...
a. Can you write the basic reason of resonance? Is there any role of natural frequency in this
phenomenon or not?
b. Generally it is accepted about the slope at any point on the displacement graph of an
object in SHM (Simple Harmonic Motion) is its velocity. Do you agree with this or not?
1.The physical reason for resonance is the transfer of energy without looses between the external applied force and the harmonic oscillator. This leads to higher and higher amplitudes of the oscillation.
For an undamped (zero external applied force) free harmonic oscillator the equation of motion is
`F = m*a` or equivalent
`-k*x = m*(d^2x)/(dt^2)` , `m*(d^2x)/(dt^2) +kx =0` , `(d^2x)/dt^2 +k/m*x =0`
and by defining the natural frequency of resonance as
`omega_0 = sqrt(k/m)` we have `(d^2x)/dt^2 + omega_0^2*x =0`
The solutions of this equation (zero driven force) are
`X(t) = X0*sin(omega_0*t + phi)`
where constants `X_0` (amplitude of oscillation) and `phi` (initial phase are to be determined from initial conditions (at `t =0`)
Now suppose we have a sinusoidal force `F = F_0*sin(omega*t)` that drives the oscillator having `omega!=omega_0` .
`m*(d^2x)/(dt^2) +k*x =F_0*sin(omega*t)`
`(d^2x)/(dt^2) +omega_0^2*x =(1/m)*F_0*sin(omega*t)`
The solutions of this equation are
`x(t) = F_0/(m*Z*omega) * sin(omega*t +Phi)`
where `Z` is an impedance
`Z = (omega_0^2-omega^2)/omega^2`
which are oscillations having same angular frequency with the applied external force.
Now suppose you have `omega=omega_0` , i.e. resonance. Because the impedance `Z = 0` in this case the amplitude of the oscillations will grow higher and higher until an infinite value. Therefore the resonance is obtained for a frequency of the external applied force equal to the natural frequency of the oscillator.
2. What you are saing about the slope of a point on the displacement graph (versus time) and the speed of SHM (at that paticular time) is true.
If we write the displacement of a SHM as a function of time as
`x(t) = A*sin(omega*t +phi)`
its slope is equal to the first derivative.
`dx/dt = omega*A*cos(omega*t+ phi)`
By DEFINITIONthe speed of a point having the displacement law `x(t)` is
`V(t)= dx/dt = omega*A*cos(omega*t +phi)`