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1 A certain mass is attached to a mass of 0.500kg. If it has a spring constant of 25.0 N/m and oscillation has an amplitude 0.15m. Calculate (a) Total energy (b)Find the potential and kinetic energy equations with respect to time

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2 parts
a

0.281 J\text{0.281 J}

b

0.281cos⁡2(7.071t) J0.281 \cos^2(7.071 t) \, \text{J}

0.281sin⁡2(7.071t) J0.281 \sin^2(7.071 t) \, \text{J}

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Here's how to solve the problem:

Part a· 2 steps

Total energy

  1. The total energy EE in a simple harmonic oscillator is given by the formula E=12kA2E = \frac{1}{2} k A^2, where kk is the spring constant and AA is the amplitude.

    Given k=25.0 N/mk = 25.0 \, \text{N/m} and A=0.15 mA = 0.15 \, \text{m}.

  2. Substitute the values into the formula and calculate the total energy.

    E=12(25.0 N/m)(0.15 m)2E = \frac{1}{2} (25.0 \, \text{N/m}) (0.15 \, \text{m})^2 E=12(25.0 N/m)(0.0225 m2)E = \frac{1}{2} (25.0 \, \text{N/m}) (0.0225 \, \text{m}^2) E=0.28125 JE = 0.28125 \, \text{J}

    The total energy is 0.281 J\boxed{\text{0.281 J}}.

Part b

0.281cos⁡2(7.071t) J0.281 \cos^2(7.071 t) \, \text{J}

Potential and kinetic energy equations with respect to time

  1. First, calculate the angular frequency ω\omega using the formula ω=km\omega = \sqrt{\frac{k}{m}}.

    Given m=0.500 kgm = 0.500 \, \text{kg} and k=25.0 N/mk = 25.0 \, \text{N/m}.

    ω=25.0 N/m0.500 kg\omega = \sqrt{\frac{25.0 \, \text{N/m}}{0.500 \, \text{kg}}} ω=50 s−2\omega = \sqrt{50 \, \text{s}^{-2}} ω≈7.071 rad/s\omega \approx 7.071 \, \text{rad/s}

  2. The potential energy U(t)U(t) of a simple harmonic oscillator is given by U(t)=12kx(t)2U(t) = \frac{1}{2} k x(t)^2.

    Assuming the oscillation starts at maximum displacement (x(t)=Acos⁡(ωt)x(t) = A \cos(\omega t)), the potential energy equation is:

    U(t)=12kA2cos⁡2(ωt)U(t) = \frac{1}{2} k A^2 \cos^2(\omega t)

    Substitute the values for kk, AA, and ω\omega:

    U(t)=12(25.0 N/m)(0.15 m)2cos⁡2(7.071t)U(t) = \frac{1}{2} (25.0 \, \text{N/m}) (0.15 \, \text{m})^2 \cos^2(7.071 t) U(t)=0.281cos⁡2(7.071t) JU(t) = \boxed{0.281 \cos^2(7.071 t) \, \text{J}}

  3. The kinetic energy K(t)K(t) of a simple harmonic oscillator is given by K(t)=12mv(t)2K(t) = \frac{1}{2} m v(t)^2.

    Using v(t)=−Aωsin⁡(ωt)v(t) = -A\omega \sin(\omega t), this simplifies to:

    K(t)=12kA2sin⁡2(ωt)K(t) = \frac{1}{2} k A^2 \sin^2(\omega t)

    Substitute the values for kk, AA, and ω\omega:

    K(t)=12(25.0 N/m)(0.15 m)2sin⁡2(7.071t)K(t) = \frac{1}{2} (25.0 \, \text{N/m}) (0.15 \, \text{m})^2 \sin^2(7.071 t) K(t)=0.281sin⁡2(7.071t) JK(t) = \boxed{0.281 \sin^2(7.071 t) \, \text{J}}
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