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Posted by on Wednesday, July 11, 2012 at 3:30pm.

A 100-g block hangs from a spring with k = 5.3 N/m. At t = 0 s, the block is 20.0 cm below the equilibrium position and moving upward with a speed of 194 cm/s. What is the block's speed when the displacement from equilibrium is 31.5 cm?

  • Physics - , Wednesday, July 11, 2012 at 3:52pm

    ω=sqrt(k/m)=sqrt(4.6/0.1) = 6.76rad/s.
    x=A•sinωt
    v=dx/dt=A•ω•cosωt

    Divide the first equation by the second and obtain
    x/v= A•sinωt/ A•ω•cosωt= ω•tanωt,
    tanωt = x•ω/v=20•6.76/194=0.7
    ωt =arctanωt =0.61 rad,
    sinωt = 0.571
    A=x/sinωt = 20/0.571=35 cm.
    For the second case:
    x1=A•sinωt1,
    sinωt1 =x1/A=31.5/35 =0.9.
    cos ωt1=sqrt(1-sin²ωt1)=0.44.
    v1= =A•ω•cosωt1=
    =35•6.76•0.44=104 cm/s.

  • Physics - , Wednesday, July 11, 2012 at 4:04pm

    K is 5.3 N/m. Why are you using 4.6 N/m?

  • Physics - , Wednesday, July 11, 2012 at 4:08pm

    This is the problem similar to that I've solved for another post. Substitute your data and calculate yourself

  • Physics - , Wednesday, July 11, 2012 at 4:13pm

    Check my calculations.
    ω=sqrt(k/m)=sqrt(5.3/0.1) = 7.28rad/s.
    x=A•sinωt
    v=dx/dt=A•ω•cosωt

    Divide the first equation by the second and obtain
    x/v= A•sinωt/ A•ω•cosωt= ω•tanωt,
    tanωt = x•ω/v=20•7.28/194=0.75
    ωt =arctanωt =0.64 rad,
    sinωt = 0.6
    A=x/sinωt = 20/0.571=33.3 cm.
    For the second case:
    x1=A•sinωt1,
    sinωt1 =x1/A=31.5/33.3 =0.95.
    cos ωt1=sqrt(1-sin²ωt1)=0.33.
    v1= =A•ω•cosωt1=
    =33.3•7.28•0.33=80 cm/s.

  • Physics - , Wednesday, July 11, 2012 at 4:22pm

    I got a similar answer of 79.0 cm/s using the conservation of energy equation.

    U(i) + K(i) = U(f) + U(f)

    Thank you for the help. And sorry for asking this question. I didn't see the similar question when I searched mine in the search bar.

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