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A-body-undergoing-SHM-about-the-origin-has-its-equation-x-0-2cos-5pit-Find-its-average-speed-from-t-0-to-t-0-7-sec-




Question Number 32585 by rahul 19 last updated on 28/Mar/18
A body undergoing SHM about the  origin has its equation x=0.2cos 5πt.  Find its average speed from t=0 to  t=0.7 sec.
AbodyundergoingSHMabouttheoriginhasitsequationx=0.2cos5πt.Finditsaveragespeedfromt=0tot=0.7sec.
Commented by rahul 19 last updated on 28/Mar/18
Commented by rahul 19 last updated on 28/Mar/18
pls explain this:  t=((nT)/4)+t_0 ..........
plsexplainthis:t=nT4+t0.
Answered by Joel578 last updated on 28/Mar/18
x(t) = 0.2 cos (5πt)  x(0) = 0.2 cos (0) = 0.2  x(0.7) = 0.2 cos ((7/2)π) = 0    v^−  = ((Δx)/(Δt)) = ((0.2)/(0.7)) = (2/7) ms^(−1)
x(t)=0.2cos(5πt)x(0)=0.2cos(0)=0.2x(0.7)=0.2cos(72π)=0v=ΔxΔt=0.20.7=27ms1
Commented by rahul 19 last updated on 28/Mar/18
ans. given is 2 m/s.
ans.givenis2m/s.
Answered by tanmay.chaudhury50@gmail.com last updated on 20/May/18
average speed=((x_2 −x_1 )/(t_2 −t_1 ))  =((0.2cos(5Π×0.7)−0.2cos(5Π×0))/(0.7−0.0))  ={0.2cos(((5×22×0.7)/7))−0.2}/0.7  =((0.2(cos11)−0.2)/(0.7))  =(2/7)×{(cos11)−1}  =(2/7)×(0.0044256−1)  =−0.2844  pls check
averagespeed=x2x1t2t1=0.2cos(5Π×0.7)0.2cos(5Π×0)0.70.0={0.2cos(5×22×0.77)0.2}/0.7=0.2(cos11)0.20.7=27×{(cos11)1}=27×(0.00442561)=0.2844plscheck
Commented by tanmay.chaudhury50@gmail.com last updated on 20/May/18
Answered by ajfour last updated on 20/May/18
Time period=((2π)/(5π))=0.4s  Distance traveled in 0.7s (=((7T)/4))  is  = 7A =7×0.2 =1.4 units          Average speed for this   duration = ((1.4units)/(0.7s)) = 2units/s .
Timeperiod=2π5π=0.4sDistancetraveledin0.7s(=7T4)is=7A=7×0.2=1.4unitsAveragespeedforthisduration=1.4units0.7s=2units/s.

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