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2 years ago

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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 t

he 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.

1 answer:

Free_Kalibri [48]2 years ago

3 0

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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