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.

X(t) = 13/13 cos(12t+α)

C =13/13

π/6 s

Explanation:

(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/s

X(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 = c

X(t) = 13/13 cos(12t+α)

since, C>0 : damped forced vibration c_1>0, c_2>0

phase angle 2π+tan^-1(c_2/c_1)

                        =2π+tan^-1(-5/12/1)= 5.884

period: T =2π/w_o

                =π/6 s

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong