A toy is undergoing SHM on the end of a horizontal spring with force constant 310 N/m . When the toy is 0.120 m from its equilibrium positio

Question

A toy is undergoing SHM on the end of a horizontal spring with force constant 310 N/m . When the toy is 0.120 m from its equilibrium position, it is observed to have a speed of 3 m/s and a total energy of 3.6 J .
1.) Find the mass of the toy.
2.) Find the amplitude of the motion.
3.) Find the maximum speed obtained by the object during its motion.

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Ngọc Khuê 6 months 2021-08-04T07:19:45+00:00 1 Answers 21 views 0

Answers ( )

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    2021-08-04T07:21:32+00:00

    Answer:

    1) the mass of the toy is 0.304 kg

    2) the amplitude of the motion is 0.1524 m

    3) the maximum speed obtained by the object during its motion is 4.87 m/s

    Explanation:

    Given the data in the question;

    force constant of spring k = 310 N/m

    position of toy x = 0.120 m

    speed v = 3 m/s

    Total energy E = 3.6 J

    let m represent mass of toy.

    1.) Find the mass of the toy.

    we know that; The total energy of the system equals the sum of kinetic energy of the toy and the potential energy of the spring;

    E = \frac{1}{2}kx² + \frac{1}{2}mv²

    we substitute

    3.6 = ( \frac{1}{2} × 310 × (0.120)² ) + ( \frac{1}{2} × m × (3)² )

    3.6 = 2.232 + 4.5m

    3.6 – 2.232 = 4.5m

    1.368 = 4.5m

    m = 1.368 / 4.5

    m = 0.304 kg

    Therefore, the mass of the toy is 0.304 kg

    2) Find the amplitude of the motion.

    we know that;

    E = \frac{1}{2}kA²

    where A is the amplitude of the motion,

    we substitute

    3.6 = \frac{1}{2} × 310 × A²

    3.6 = 155 × A²

    A² = 3.6 / 155

    A² = 0.0232258

    A = √0.0232258

    A = 0.1524 m

    Therefore, the amplitude of the motion is 0.1524 m

    3) Find the maximum speed obtained by the object during its motion;

    we know that;

    E = \frac{1}{2}mv_{max²

    where v_{max is the maximum speed

    so we substitute

    3.6 = \frac{1}{2} × 0.304 × v_{max²

    3.6 = 0.152 × v_{max²

    v_{max² = 3.6 / 0.152

    v_{max² = 23.6842

    v_{max² = √23.6842

    v_{max² = 4.8664 ≈ 4.87 m/s

    Therefore, the maximum speed obtained by the object during its motion is 4.87 m/s

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