? asked in 教育與參考考試 · 1 decade ago

物理好難阿 請懂的進來替看看

1 An apple weighs 1.00 N. When you hang it from the end of a long spring of force constant 1.50 N/m and negligible mass, it bounces up and down in SHM. If you stop the bouncing and let the apple swing from side to side through a small angle, the frequency of this simple pendulum is half the bounce frequency. (Because the angle is small, the back-and-forth swings do not cause any appreciable change in the length of the spring.) What is the unstretched length of the spring (with the apple removed)?



The two pendulums shown In Fig. 13.34 each consist of a uniform solid ball of mass M supported by a massless string, but the ball for pendulum B is much larger. Find the period of each pendulum for small displacements. Which ball takeslonger to complete a swing?

2 Answers

  • Anonymous
    1 decade ago
    Favorite Answer

    For the first case (mass and spring system),

    Angular frequency, ω = √k/m

    As weight = mg

    Take g = 10 ms^-2

    So, 1.00 = 10m

    Mass, m = 0.10 kg

    So, angular frequency

    ω = √1.5/0.10

    ω = √15

    frequency, f = ω/2π = √15 / 2π Hz

    or the second case (simple pendulum),

    Angular frequency, ω’ = √g/l = √10/l

    Frequency, f’ =ω’/2π

    Take g = 10 ms^-2 and f = 2f’

    So, √15 / 2π= √10/l / 2π

    15 = 10/l

    unstretched length, l = 0.667 m

    For question 2, as you didn't provide the graph, so I can't help you solve the problem.

    Source(s): Myself~~~
  • Anonymous
    7 years ago

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