Question

A block of mass m is undergoing SHM on a horizontal, frictionless surface while attached to a light, horizontal spring. The spring has force constant k, and the amplitude of the motion of the block is A.

1) The average speed is the total distance traveled by the block divided by the time it takes it to travel this distance. Calculate the average speed for one cycle of the SHM.

Express your answer in terms of the variables m, A, and k.

2) How does the average speed for one cycle compare to the maximum speed vmax?  

3) Is the average speed more or less than half the maximum speed? Based on your answer, does the block spend more time while traveling at speeds greater than vmax/2 or less than vmax/2?

Answer 1

Answer 2

Answer 3

1) Average Speed for One Cycle:

• Total distance per cycle (SHM):
  $4A$ (moves $A\to -A\to A$).

• Period ($T$) of SHM:
  $T=2\pi \sqrt{\frac{m}{k}}$.

• Average speed ($v_{\text{avg}}$):

  $$v_{\text{avg}}=\frac{\text{Total distance}}{T}=\frac{4A}{2\pi \sqrt{\frac{m}{k}}}=\frac{2A}{\pi }\sqrt{\frac{k}{m}}\mathrm{.}$$

Answer:

$$v_{\text{avg}}=\boxed{{\displaystyle \frac{2A}{\pi }\sqrt{\frac{k}{m}}}}$$

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2) Comparison with Maximum Speed ($v_{\text{max}}$):

• Maximum speed:
  $v_{\text{max}}=A\sqrt{\frac{k}{m}}$ (at equilibrium).

• Ratio:

  $$\frac{v_{\text{avg}}}{v_{\text{max}}}=\frac{\frac{2A}{\pi }\sqrt{\frac{k}{m}}}{A\sqrt{\frac{k}{m}}}=\frac{2}{\pi }\approx \mathrm{0.637.}$$

Answer:
Average speed is ${\displaystyle \frac{2}{\pi }}$ (≈63.7%) of $v_{\text{max}}$.

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3) Time Spent Above/Below $v_{\text{max}}\mathrm{/}2$:

• Since $v_{\text{avg}}>{\displaystyle \frac{v_{\text{max}}}{2}}$:
  The block spends more time at speeds less than ${\displaystyle \frac{v_{\text{max}}}{2}}$ (because high speeds occur briefly near equilibrium).

• 
  

Answer:

• Average speed is more than half of $v_{\text{max}}$.

• The block spends more time at speeds < ${\displaystyle \frac{v_{\text{max}}}{2}}$.