A cart of mass 350 g is placed on a frictionless horizontal air track. A spring having a spring constant of 7.5 N/m is attached between the

Question

A cart of mass 350 g is placed on a frictionless horizontal air track. A spring having a spring constant of 7.5 N/m is attached between the cart and the left end of the track. The cart is displaced 3.8 cm from its equilibrium position. (a) Find the period at which it oscillates. s (b) Find its maximum speed. m/s (c) Find its speed when it is located 2.0 cm from its equilibrium position.

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Doris 3 years 2021-08-23T08:38:44+00:00 1 Answers 11 views 0

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  1. Nem
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    2021-08-23T08:39:59+00:00
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    Answer:

    (a) T = 1.35 s

    (b) vmax = 0.17 m/s

    (c) v = 0.056 m/s

    Explanation:

    (a) In order to calculate the period of oscillation you use the following formula for the period in a simple harmonic motion:

    T=2\pi\sqrt{\frac{m}{k}}          (1)

    m: mass of the cart = 350 g = 0.350kg

    k: spring constant = 7.5 N/m

    T=2\pi \sqrt{\frac{0.350kg}{7.5N/m}}=1.35s

    The period of oscillation of the car is 1.35s

    (b) The maximum speed of the car is given by the following formula:

    v_{max}=\omega A       (2)

    w: angular frequency

    A: amplitude of the motion = 3.8 cm = 0.038m

    You calculate the angular frequency:

    \omega=\frac{2\pi}{T}=\frac{2\pi}{1.35s}=4.65\frac{rad}{s}              

    Then, you use the result of w in the equation (2):

    v_{max}=(4.65rad/s)(0.038m)=0.17\frac{m}{s}

    The maximum speed if 0.17m/s

    (c) To find the speed when the car is at x=2.0cm you first calculate the time t by using the following formula:

    x=Acos(\omega t)\\\\t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\t=\frac{1}{4.65rad/s}cos^{-1}(\frac{0.02}{0.038})=0.069s

    The speed is the value of the following function for t = 0.069s

    |v|=|\omega A sin(\omega t)|\\\\|v|=(4.65rad/s)(0.038m)sin(4.65rad/s (0.069s))=0.056\frac{m}{s}

    The speed of the car is 0.056m/s

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