application of simple harmonic oscillator for a woodpecker Confused please help!!?

Question

You observe a woodpecker in a tree, and decide to model the motion of its head as it taps for insects as a simple harmonic oscillator.
You measure a frequency of 8.0 taps per second, an amplitude of the motion of the woodpecker's beak of 1.7 cm, and you estimate the mass of the woodpecker's head is 53 grams.

(a) what is the maximum acceleration that the head and beak undergo as the woodpecker pecks?Express this in both MKS units, and in units of 'g' (9.8 m/s^2)

(b) what is the maximum speed of the woodpeckers head during the motion?

thank you very much... :)

Andrew Smith · Answer

for SHM a max = w^2 R  = (2pi() f)^2 * r  
Now the first question is "What is R" .  When you say that the amplitude is 1.7cm I have to assume that this is the whole distance from one end to the other so that r = 0.017/ 2 m

therefore a =  (2 * pi() * 8)^2 * 0.017/2  ~=21.5 m/s^2

The maximum speed is wr = 2 * pi() * 8 * 0.017/2  = 0.43 m/s

Note that the key difference between this and nyphdinmd's answer is that he has assumed that R = 0.017 m and I have assumed that the whole motion is 0.017m  otherwise we agree on our answers.

nyphdinmd · Answer

If you assume the beak moves according to  y = A*cos(wt) where A = amplitude, and w = 2*pi* frequency  (f), then speed and acceleration are given by

v = dy/dt = -A*w*sin(wt)

a = d^2y/dt^2 = - A*w^2*cos(wt)  -->  max accel. occurs when cos(wt) = -1 so a_max = A*w^2
435.2*pi
  w = 2*pi*f = 2*pi*8./sec = 16*pi rad/s  -->  a_max = 1.7cm*(16*pi*rad/s)^2 = 435.2 *pi cm/s^2 = 4.325*pi m/s^2

g = 980 cm/s^2 --->  a_max = 0.444*pi g = 1.395 g

Make speed occurs when sin(wt) = -1 --->  v_max = A*w = 0.272*pi m/s = 0.85 m/s

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application of simple harmonic oscillator for a woodpecker Confused please help!!?

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