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sergij07 [2.7K]
3 years ago
11

A spring with a mass of 5 Kg has natural length 0.5m. A force of 35.6 N is required to maintain it stretched to a length of 0.5m

. If the spring is stretched to a length of 0.5m and released with initial velocity 0, find the position of the mass at any time t. Here damping constant is zero.
Physics
1 answer:
aleksandrvk [35]3 years ago
8 0

Answer:

Explanation:

force constant of spring  k = force / extension

= 35.6 / 0.5

k = 71.2 N / m

angular frequency ω of oscillation by spring mass system

\omega = \sqrt{\frac{k}{m} }

where m is mass of the body attached with spring

Putting the values

\omega = \sqrt{\frac{71.2}{5} }

ω = 3.77 radian / s

The oscillation of the mass will be like SHM having amplitude of 0.5 m and angular frequency of 3.77 radian /s . Initial phase will be π / 2

so the equation for displacement from equilibrium position that is middle point can be given as follows

x = .5 sin ( ω t + π / 2 )

= 0.5 cos ω t

= 0.5 cos 3.77 t .

x = 0.5 cos 3.77 t .

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a) The position is requested for several times

X axis

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          vₓ = x / t

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c) The launch angle      θ = 33.7ºto)

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