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## A body with mass of 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time tequals0 the body is pulled

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A body with mass of 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time tequals0 the body is pulled 1 m to the left, stretching the spring, and set in motion with an initial velocity of 20 m/s to the left. (a) Find x(t) in the form Upper C cosine (omega 0 t minus alpha ). (b) Find the amplitude and the period of motion of the body.

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Physics
3 years
2021-07-14T07:49:53+00:00
2021-07-14T07:49:53+00:00 1 Answers
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## Answers ( )

Answer:X(t) = 13/13 cos(12t+α)C =13/13π/6 sExplanation:(A) A body with mass 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time t = 0 the body is pulled 1 m in to the right, stretching the spring, 3 set in motion with an initial velocity of 5 m/s to the left.

(a) Find X(t) in the form c • cos(w_o*t— α)

(b) Find the amplitude 3 Period of motion of the body 1

mass: m = 200g = 0.200 kg

displacement: ΔX = 20 cm = 0.20 m

Spring Constant: K = 9/0.20 = 45 N/m

IV: X(0) = 1m V(0) = -5 m/s

Simple Harmonic Motion: c•cos(cosw_t— α) = X(t)Circular Frequency: w_o= √k/m= √36/(0.20) = 13 rad/sX(0) = 1m =c_1

X'(0) = V(0) =

c_2*w_o/w_o

= -5/12 =

c_2

“radians Technically Unitless”

Amplitude: c= √ci^2 + c^2 ==> √1^2 + (-5/12)^2 = 1 m =13/13 = cX(t) = 13/13 cos(12t+α)since, C>0 : damped forced vibrationc_1>0, c_2>0phase angle2π+tan^-1(c_2/c_1)=2π+tan^-1(-5/12/1)= 5.884

period: T =2π/w_o=π/6 s