Solve Oscillation Problem: |v|=0.5v_max

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SUMMARY

The discussion focuses on solving the oscillation problem involving a block attached to a spring, initially displaced to x = +4.9 m with a period of 0.5 s. The condition |v| = 0.5 v_max is analyzed, where v_max is determined to be 9.8π. The velocity function is derived as v(t) = 19.6π cos(4πt + 0.5π), leading to the conclusion that |v| equals half of v_max when the cosine function equals 1/2, specifically at the angles π/3 and 5π/3. The positions and times corresponding to these conditions are to be calculated based on the derived equations.

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Homework Statement


A block attached to a spring is displaced from equilibrium to the position x = +4.9 m and released. The period is 0.5 s . At what positions and times during the first complete cycle do the following condition occur:

| v | = 0.5 v_{max}, Where v_{max} is the maximum speed?


Homework Equations


none i suppose.


The Attempt at a Solution


I believe the s(t)= 4.9 sin (4pi(t)+0.5pi)
v(t) = 19.6pi cos (4pi(t)+0.5pi)

there for 0.5 max velocity is = 9.8pi ---> (19.6pi/2)

so for velocity without breaking the phase constant:

4pi (t) + 0.5pi = pi/3 5pi/3

i don't think i am right, can anyone clear it up?
 
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Hi phat2107! :wink:

Yes, that looks ok … |v| is half vmax when |cos| = 1/2 …

so that's when t and s = … ? :smile:​
 

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