An air-track glider attached to a spring oscillates between the 12.0 cm mark and the 65.0 cm mark on the track. The glider completes 14.0 os

Question

An air-track glider attached to a spring oscillates between the 12.0 cm mark and the 65.0 cm mark on the track. The glider completes 14.0 oscillations in 34.0 s . What are the (a) period, (b) frequency, (c) amplitude, and (d) maximum speed of the glider

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Dâu 4 years 2021-07-18T12:37:17+00:00 2 Answers 15 views 0

Answers ( )

  1. taiduc
    0
    2021-07-18T12:38:27+00:00
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    Answer:

    a) T = 2.429 s

    b) f = 0.412 Hz

    c) A = 26.5 cm

    d) v = 68.6 cm/s

    Explanation:

    a) The period is equal:

    T = 34/14 = 2.429 s

    b) The frequency is equal:

    f = 1/T = 1/2.429 = 0.412 Hz

    c) The amplitude is equal:

    A=\frac{62-12}{2} =26.5cm

    d) The maximum speed of the glider is:

    v=aw=A*2\pi *f=26.5*2*\pi *0.412=68.6cm/s

  2. huyenthanh
    0
    2021-07-18T12:38:48+00:00
    Reply

    Answer:

    (a) 2.4s

    (b) 0.41Hz

    (c) 26.5cm

    (d) 68.1m/s

    Explanation:

    (a) To calculate the period we can use the formula:

    T=\frac{1}{f}=\frac{1}{\frac{14.0}{34.0s}}=2.4s

    (b) the frequency is the inverse of period:

    f=\frac{1}{T}=\frac{1}{2.4}=0.41Hz

    (c) The amplitude is the max distance to the equilibrium position, that is:

    A=\frac{65.0cm-12.0cm}{2}=26.5cm

    (d) the maximum speed of the glider can be computed by using:

    v_{max}=\omega A\\\\\omega=2\pi f=2\pi (0.41Hz)=2.57rad/s\\\\v_{max}=(2.57rad/s)(26.5cm)=68.105\frac{cm}{s}

    hope this helps!!

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