A)
B)
C)
D)
Correct Answer: D
Solution :
[d] y = A Sin \[\omega \,t\] \[\therefore \]We do not use y = A Cos \[\omega \,t\]because as per as graph cos function does not satisfy. But Sin function start from origin as shown in graph. \[v=A\,\omega \,Cos\,\omega t\,\therefore \left[ v=\frac{dy}{dt} \right]\] \[F\propto a\] Graph starts from origin (Sin function starts from origin) but with -ve amplitude
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