? 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)?

2

Update:

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

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  • 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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