please ans the fast ..........,........
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(i) displasement is x = 5 sin (2π/T t ) = 5 sin (π t ) [cm] x(t=1/3) = 5 sin (π/3) = 5*√3/2.
(ii) then the velosity would be v = 5π cos (π t) [cm/s] v(t=1/3) =2.5π
(iii) acceleration a = -5 π2 sin (π t) a(t=1/3)= -2.5π2√3
bro i dont understand
in this question we used standerd eq of shm
Time period t=2s
Amplitude a=5cm
Time after which displacement is to be calculated is t=1/3
At that time a=0
Amplitude a=5cm
Time after which displacement is to be calculated is t=1/3
At that time a=0
X=asin(wt+a)
X=5sin(¶×1/3+0)
X=5sin(¶/3)
X=5*sin60
X=5*(3/2)^1/2
X=8.66/2
X=4.33
X=4cm approx
X=5sin(¶×1/3+0)
X=5sin(¶/3)
X=5*sin60
X=5*(3/2)^1/2
X=8.66/2
X=4.33
X=4cm approx
Similar questions
Amplitude a=5cm
Time after which displacement is to be calculated is t=1/3
At that time a=0
X=asin(wt+a)
X=5sin(¶×1/3+0)
X=5sin(¶/3)
X=5*sin60
X=5*(3/2)^1/2
X=8.66/2
X=4.33
X=4cm approx