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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-

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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.
$$\boldsymbol{{A}}\:{body}\:{undergoing}\:{SHM}\:{about}\:{the} \\ $$$${origin}\:{has}\:{its}\:{equation}\:{x}=\mathrm{0}.\mathrm{2cos}\:\mathrm{5}\pi{t}. \\ $$$${Find}\:{its}\:{average}\:{speed}\:{from}\:{t}=\mathrm{0}\:{to} \\ $$$${t}=\mathrm{0}.\mathrm{7}\:{sec}. \\ $$
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 ..........
$${pls}\:{explain}\:{this}: \\ $$$${t}=\frac{{nT}}{\mathrm{4}}+{t}_{\mathrm{0}} ………. \\ $$
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)
$${x}\left({t}\right)\:=\:\mathrm{0}.\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{5}\pi{t}\right) \\ $$$${x}\left(\mathrm{0}\right)\:=\:\mathrm{0}.\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{0}\right)\:=\:\mathrm{0}.\mathrm{2} \\ $$$${x}\left(\mathrm{0}.\mathrm{7}\right)\:=\:\mathrm{0}.\mathrm{2}\:\mathrm{cos}\:\left(\frac{\mathrm{7}}{\mathrm{2}}\pi\right)\:=\:\mathrm{0} \\ $$$$ \\ $$$$\overset{−} {{v}}\:=\:\frac{\Delta{x}}{\Delta{t}}\:=\:\frac{\mathrm{0}.\mathrm{2}}{\mathrm{0}.\mathrm{7}}\:=\:\frac{\mathrm{2}}{\mathrm{7}}\:\mathrm{ms}^{−\mathrm{1}} \\ $$
Commented by rahul 19 last updated on 28/Mar/18
ans. given is 2 m/s.
$${ans}.\:{given}\:{is}\:\mathrm{2}\:{m}/{s}. \\ $$
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
$${average}\:{speed}=\frac{{x}_{\mathrm{2}} −{x}_{\mathrm{1}} }{{t}_{\mathrm{2}} −{t}_{\mathrm{1}} } \\ $$$$=\frac{\mathrm{0}.\mathrm{2}{cos}\left(\mathrm{5}\Pi×\mathrm{0}.\mathrm{7}\right)−\mathrm{0}.\mathrm{2}{cos}\left(\mathrm{5}\Pi×\mathrm{0}\right)}{\mathrm{0}.\mathrm{7}−\mathrm{0}.\mathrm{0}} \\ $$$$=\left\{\mathrm{0}.\mathrm{2}{cos}\left(\frac{\mathrm{5}×\mathrm{22}×\mathrm{0}.\mathrm{7}}{\mathrm{7}}\right)−\mathrm{0}.\mathrm{2}\right\}/\mathrm{0}.\mathrm{7} \\ $$$$=\frac{\mathrm{0}.\mathrm{2}\left({cos}\mathrm{11}\right)−\mathrm{0}.\mathrm{2}}{\mathrm{0}.\mathrm{7}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{7}}×\left\{\left({cos}\mathrm{11}\right)−\mathrm{1}\right\} \\ $$$$=\frac{\mathrm{2}}{\mathrm{7}}×\left(\mathrm{0}.\mathrm{0044256}−\mathrm{1}\right) \\ $$$$=−\mathrm{0}.\mathrm{2844} \\ $$$${pls}\:{check} \\ $$
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 .
$${Time}\:{period}=\frac{\mathrm{2}\pi}{\mathrm{5}\pi}=\mathrm{0}.\mathrm{4}{s} \\ $$$${Distance}\:{traveled}\:{in}\:\mathrm{0}.\mathrm{7}{s}\:\left(=\frac{\mathrm{7}{T}}{\mathrm{4}}\right) \\ $$$${is}\:\:=\:\mathrm{7}{A}\:=\mathrm{7}×\mathrm{0}.\mathrm{2}\:=\mathrm{1}.\mathrm{4}\:{units} \\ $$$$\:\:\:\:\:\:\:\:{Average}\:{speed}\:{for}\:{this}\: \\ $$$${duration}\:=\:\frac{\mathrm{1}.\mathrm{4}{units}}{\mathrm{0}.\mathrm{7}{s}}\:=\:\mathrm{2}{units}/{s}\:. \\ $$

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