A body oscillates with simple harmonic motion along the x-axis. its displacement varies with time according to the equation x(t)
= a sin(ω t + φ). if a = 5 m, ω = 3.444 rad/s, and φ = 1.0472 rad, what is the acceleration of the body at t = 3 s? note: the argument of the sine function is in radians rather than degrees. answer in units of m/s 2 .
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Answer:
x=4
Step-by-step explanation:
<u>Step 1</u>:-
given the lengths of two sides of a right angle are represented by 2x and 3(x+1) and longest side is 17 units.
AB = 2x and BC = 3(x+1) and longest side AC= 17
by using Pythagoras theorem
<u>step 2:-</u>
The hypotenuse is longest side is AC = 17 units
(17)^2 = 4x^2 +9(X+1)^2
on simplification, we will use formula
289 = 4x^2 +9(x^2+2x+1)
finding factors 70 X 52 = 3640
Taking common , we get
13x(x-4)+70(x-4)=0
x-4=0 and 13x+70=0
x=4 and
x=4 and
we can not choose negative value so x value is 4
Final answer:- x = 4
<u>verification:-</u>
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289 = 4(4)^2+9(4+1)^2
289 = 64 +9(25)
289=289