a) What happens to the graph if the amplitude is doubled and the frequency is ha

Question

a) What happens to the graph if the amplitude is doubled and the frequency is halved?

b) What happens to the graph if the amplitude and spring constant are kept the same but the mass is quadrupled?

An object oscillating on a spring has the velocity graph shown in the figure. Parts A and B are independent questions, each starting from the graph shown. Draw a velocity graph if the following changes are made. a) What happens to the graph if the amplitude is doubled and the frequency is halved? b) What happens to the graph if the amplitude and spring constant are kept the same but the mass is quadrupled?

Explanation / Answer

If you look at the equation of velocity for a simple harmonic oscillator (a spring in this case) it has the form v(t) = -Acos(ut) where A is the amplitude, and u is the angular frequency. Increasing the amplitude only changes how high up or down the graph moves, but not how long it takes to complete a cycle. So doubling the amplitude of this graph would cause the peaks and troughs to appear at 4 and -4 instead of 2 and -2, but this alone will not change where it crosses the x-axis or how far it goes before completing a cycle. Halving the frequency, on the other hand, does the opposite. It doesn't change how far up or down on the y-axis thsi graph extends, but it does change how quickly the object oscillates, i.e. it will change how quickly the graph completes a cycle. One cycle of the graph is given by the following formula 1 Cycle (Period) = 2pi/u Keep in mind though, that we are told that the frequency is halved, not the angular frequency. The angular frequency (u) is equal to 2pi times the frequency i.e. angular frequency (u) = 2*pi*frequency(f) = 2*pi/cycle period (T) While these relationships are useful in general, for this particular case they simplify nicely to the relationship: Frequency = 1/Cycle period So, doubling the frequency will halve the cycle period so this graph will complete a cycle twice as fast. So, doubling the frequency and doubling the ampliutude will make this graph look "squished" it will extend higher up and down on the y-axis, and will complete cycles faster. For the second problem, we can use the relationship that can be derived from the equations of motion for s simple harmonic oscillator: Angular frequency (u) = sqrt(spring constant(k) / mass (m)) so, if the spring constant is left alone but the mass is quadrupled, then we get the following relationship: Let u = original angular frequency Let u2 = angular frequency after quadrupling mass k is the constant spring constant m is the original mass So the original angular frequency u = sqrt(k/m) The second angular freqeuncy u2 = sqrt(k/4m) you can pull the 1/4 out of the square root to get u2 = 1/2 sqrt(k/m) = 1/2 u original So, you can see that quadrupling the mass causes the angular frequency to decrease by a factor of two, so the object will oscillate more slowly which will cause the graph to complete cycles twice as slowly which will make the graph look stretched horizonally. Again, the amplitude doesn't change so it will still extend to 2 and -2 on the y axis.

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