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

Question

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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Ladonna 3 years 2021-07-14T07:49:53+00:00 1 Answers 29 views 0

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  1. trungdung22
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    2021-07-14T07:51:04+00:00
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    Answer:

    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

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