What is the ratio of t1 to t2 for a particle executing SHM?

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Homework Statement


For a Particle executing SHM, the displacement x of the particle as a function of time is given as x=A sin 2 pi t, where x is in cm and t is in seconds. Let the time taken by the particle to travel from x = 0 to x =A/2 be t1 and the time taken to travel from x=A/2 to x = A be t2.
Find the ration of t1/t2

Homework Equations

The Attempt at a Solution


t1/t2...

t1 = (x * 2)/(A sin 2 pi)
t2 = (x *2) /(A sin 2 pi)

as Total distance is A...
In t1 cover A/2 ( Half distance)
So in t2 also cover A/2 ( remaining half)[/B]
 
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Please HELP me to understand concept rather solution...
 
it will be at A/2 when sin(2pit)=1/2 so 2pit=pi/6 (unit circle)

so what does t equal at that time, what does 2pi t have to equal for x to be A? what does t equal to make these true (just algebra there)

relevant equations
x=Asin(ωt) as you stated ω=2pi

Δt=t(final)-t(initial)

the concept is more quantitive than qualitive, the point here to notice is that even though velocity acceleration and position are always changing, the fack that F=-kx with shm allows you to derrive an amazing amount of information based on the trig patterns they follow and those patterns simple relation to their derivatives. The range of quantitive information you can gather from so little given using the trig functions is the concept
 
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