A frictionless spring with a 3-kg mass can be held stretched 0.8 meters beyond its natural length by a force of 40 newtons. If the spring be

Question

A frictionless spring with a 3-kg mass can be held stretched 0.8 meters beyond its natural length by a force of 40 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 1 m/sec, find the position of the mass after tt seconds.

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Thiên Ân 3 years 2021-08-06T00:25:26+00:00 1 Answers 12 views 0

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  1. Sigrid
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    2021-08-06T00:26:30+00:00
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    Answer:

    Explanation:

    mass m = 3 kg

    spring constant be k

    k x .8 = 40 N

    k = 40 / .8 = 50 N /m

    angular frequency ω = √ ( k / m )

    = √ ( 50 / 3 )

    = 4.08 rad /s

    Let amplitude of oscillation be A .

    1/2 k A² = 1/2 m v²

    50 A² = 3 x 1²

    A = .245 m = 24.5 cm

    For displacement , the equation of SHM is

    x = A sinωt

    = 24.5 sin4.08 t

    x = 24.5 sin4.08 t

    Here, angle 4.08 t is in radians .

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