Physics

Solution:

The given equation can be written as

$y=3sin100πt+(4−4cos100πt)−6$

$=3sin100πt+4sin(100πt+π/2)−2$

or $y=5sin(100π−53_{∘})−2$

$y_{max}=5−2=3$

$y_{min}=−5−2=−7$

Mean position$=2y_{max}+y_{min} =−2$ cm

Solution:

$y_{1}=sin(ωt)sin(6π )+cos$ $ωtsin3π $

$=sin$ $ωtcos(3π )+cos$ $ωtsin3π $

$=sin(ωt+3π )$

$y_{2}=sin(ωt+6π )$

Thus phase difference is

$3π −6π =6π $

Solution:

$x=A$sin( $ωt+ϕ$)

$ω=202π =202π =10π secrad $

let at $t=0,x=4$, thus

$4=5$ sin $ϕ$

$sin_{−1}54 =ϕ$

Now for at $t=t_{1}$, let $x=−3$, thus we have

$5−3 =sin(ωt_{1}+ϕ)$

or

$sin_{−1}(5−3 )−sin_{−1}(54 )=ωt_{1}$

$10π(−0.643−0.927) =t_{1}$

Solving we get $t_{1}=5sec$

$a=−w_{2}x$

Any general equation satisfying the above criterion represents a simple harmonic motion.

i.e. $x=Asinwt$

$τ=−kθ$

General equation for angular displacement:

$θ=θ_{o}sin(wt+ϕ)$

$x_{1}(t)=Asin(ωt)$

$x_{2}(t)=Asin(ωt+π/4)$