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Posted by on Friday, March 30, 2012 at 1:55pm.

Objects of equal mass are oscillating up and down in simple harmonic motion on two different vertical springs. The spring constant of spring 1 is 186 N/m. The motion of the object on spring 1 has 3 times the amplitude as the motion of the object on spring 2. The magnitude of the maximum velocity is the same in each case. Find the spring constant of spring 2.

My work...

Aw=Aw
A(sqrt of k/m) = A(sqrt k/m)
(sqrt of 186) = 3 (sqrt of k)
(sqrt of 186)/3 = (sqrt of k)
[(sqrt of 186)/3] ^2 = k
I got 20.6 but this is wrong, Please help!

  • physics - , Friday, March 30, 2012 at 3:31pm

    body 1
    x = sin 2 pi f t
    v = 2 pi f cos 2 pi f t
    a = -(2pi)^2 f^2 sin 2 pi f t
    -k x = m a
    - k x = - m(2pi)^2 f^2 sin 2 pi f t
    - k/m = - (2 pi)^2 f^2
    f = (1/2pi) sqrt(k/m)

    body 2
    x = 3 sin 2 pi F t
    v = 3 (2 pi) F cos 2 pi F t
    - K/m = -3 (2 pi)^2 F^2 sin 2 pi F t
    -K/m =-(2 pi)^2 F^2
    F = (1/2pi) sqrt(K/m)

    the to v values are equal in magnitude
    2 pi f = 6 pi F

    2 pi ((1/2pi) sqrt(k/m)= 6 pi(1/2pi) sqrt(K/m)

    sqrt(k/m) = 3 sqrt(K/m)
    k/m = 9 K/m
    K = k/9 = 20.67

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