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? asked in Science & MathematicsPhysics · 6 years ago

Can I see actual steps for this, not just an answer. I would actually like to learn. thank you!?

A spring is compressed by 0.0664 m and is used to launch an object horizontally with a speed of 1.64 m/s. If the object were attached to the spring, at what angular frequency (in rad/s) would it oscillate?

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  • 6 years ago
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    k = spring constant, x = initial compression, U = potential energy, K = kinetic energy, v = final speed, m = mass, ω = angular frequency

    The potential energy of a spring U = 0.5*k*x^2. The kinetic energy of an object is K = 0.5*m*v^2. The angular frequency of a spring ω = sqrt(k/m).

    When the spring is compressed and at rest, all of the system's energy is in the spring's potential energy (E(initial) = U(initial)). When the object gets launched, all of the system's energy is in the object's kinetic energy (E(final) = K(final)). Through the conservation of energy, we know that E(initial) = E(final). So:

    U(initial) = K(final)

    0.5*k*x^2 = 0.5*m*v^2

    k*x^2 = m*v^2

    The tricky part is that we don't know m or k, so we cannot find either. However, we're only trying to find ω = sqrt(k/m). So apply some algebra as shown below:

    k*x^2 / m = v^2

    k / m = v^2 / x^2

    sqrt(k / m) = v / x

    ω = v / x

    And that's the answer.

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