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A particle executes SHM with a time period of 2s and amplitude 5 cm. Find (i) displacement (ii) velocity (iii) acceleration, after `1//3` second, starting from mean position. |
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Answer» Here, `T=2s,a=5cm,t=(1)/(3)s`. (a) Displacement , `y=asin((2pi)/(T))t=5sin((2pi)/(2))xx(1)/(3)=5xx(sqrt(3)/(2))=4.33cm` `(b)` Velocity , `V=(dy)/(dt)=a(2pi)/(T)cos((2pi)/(T))t=5xx(2pi)/(2)xos((2pi)/(2))xx(1)/(3)=7.86cm//s` `(c)` Acceleration, `A=(dV)/(dt)=-(4ppir^(2))/(T^(2))asin((2pi)/(T))t=(4xxpi^(2))/(4)xx5xxsin((2pi)/(2))xx(1)/(3)` `=(4)/(4)xx((22)/(7))^(2)xx5xx(sqrt(3))/(2)=-42.77cm//s^(2)` |
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