A 0.600-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 12.2 cm. (A

Question

A 0.600-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 12.2 cm. (Assume the position of the object is at the origin at t = 0.)

a. Calculate the maximum value of its speed
b. Calculate the maximum value of its acceleration.
c. Calculate the value of its speed when the object is 8.60 cm from the equilibrium position.
d. Calculate the value of its acceleration when the object is 8.60 cm from the equilibrium position.
e. Calculate the time interval required for the object to move from x = 0 to x = 6.60 cm.

in progress 0
Khánh Gia 4 years 2021-08-22T20:44:57+00:00 1 Answers 30 views 0

Answers ( )

  1. thaiduong
    0
    2021-08-22T20:46:39+00:00
    Reply

    Answer:

    a.   vmax = 0.445 m/s

    b.   amx = 1.65 m/s^2

    c.    v = 0.32 m/s

    d.    1.12 m/s^2

    e.     t = 0.21 s

    Explanation:

    a. The maximum speed of the object is given by the following formula:

    v_{max}=\omega A    (1)

    w: angular frequency of the object

    A: amplitude of the motion = 12.2cm = 0.122m

    The angular frequency is calculated by using:

    \omega=\sqrt{\frac{k}{m}}     (2)

    k: spring constant = 8.00 N/m

    m: mass of the object = 0.600kg

    you replace the equation (2) into the equation (1) and replace the values of the other parameters:

    v_{max}=\sqrt{\frac{k}{m}}A=(\sqrt{\frac{8.00N/m}{0.600kg}})(0.122m)=0.445\frac{m}{s}

    The maximum speed is 0.445m/s

    b. The maximum acceleration is:

    a_{max}=\omega^2 A\\\omega=\sqrt{\frac{8.00N/m}{0.600kg}}=3.65\frac{rad}{s}\\\\a_{max}=(3.65rad/s)^2(0.122m)=1.625\frac{m}{s^2}

    The maximum acceleration is 1.65m/s^2

    c. To calculate the value of the speed for x = 8.60cm you first find the time t by using the following equation of motion for a simple harmonic motion:

    x=Asin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{x}{A})\\\\t=\frac{1}{3.65rad/s}sin^{-1}(\frac{8.60cm}{12.2cm})=0.21s

    You use this value of t in the following equation for v:

    v=\omega Acos(\omega t)\\\\v=(3.65rad/s)(0.122m)cos((3.65rad/s)(0.21s))=0.32\frac{m}{s}

    The speed of the object when it is at x = 8.60cm is 0.32m/s

    d. The acceleration is:

    a=-\omega^2 A sin(\omega t)\\\\a=-(3.65rad/s)^2(0.122m)sin((3.65rad/s)(0.21))=1.12\frac{m}{s^2}

    The acceleration is 1.12 m/s^2

    e. The time is 0.21s

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )