In the May 2015 TZ1 IB Physics Paper 1 SL and HL there was a question (13SL,10HL) on the effect of damping on an oscillating system experiencing a driving force of variable frequency. The IB Examination Report states that "as damping is increased and friction is introduced into the system so the time per oscillation will increase, answer is B". This may not be intuitively obvious so a mathematical route to the solution may be helpful.
Let the frequency of vibration of the system when no damping occurs (usually just called the natural frequency) be F. The frequency of the driving force that acts on the system is f. The amplitude of the forced oscillations (this is the greatest displacement of the system from its equilibrium position) caused by the action of the driving force is A. Damping means that a resistive force is acting on the system as it vibrates causing its energy to decrease.
Driving force with no damping. When damping is not present a huge vibration known as resonance ocurs when the frequency of the disturbing force equals the natural frequency of vibration. As the amplitude is huge we can predict that the amplitude of oscillation must be inversely proportional to the difference in the frequencies. When f - F approaches zero A approaches infinity. A mathematical solution of the differential equation describing the system without damping gives A = 1/(f2-F2). How can a catastrophic vibration be avoided when f = F? Including a damping term in the equations removes some energy from the system allowing it to be unavailable to participate in a large vibration.
Driving force with damping. When a damping term is present the differential equation describing the system contains an additional term 2kv, where k is the damping coefficient and v is the velocity of the mass. The solution for A with damping is A = 1/√((f2-F2)2+ 4k2f2 ). Notice the additional term in the denominator. When f = F we have A = 1/(4k2f2) giving a finite amount of vibration when f = F. Catastrophe is avoided! Let us examine the equation
A = 1/√( (f2-F2)2+ 4k2f2 )
If we plot A versus f we obtain the graph shown on the 2015 examination paper. By differentiating A with respect to f we find that A has a maximum value of A0 when f has the value f0
f0 = √( F2-2k2 )
A0 =1/( 2k √(F2-k2) )
To answer the question, as k increases A0 decreases and f0 decreases so alternative B is the correct answer.