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A force of 300 N is used to stretch a horizontal spring with a 0.5 kg block. The length of spring is 0.25 m. The block is released from rest and undergoes simple harmonic motion. Calculate the velocity when the block is stretched 15 cm.

User Madewulf
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1 Answer

3 votes

Answer:

v = 11.34 m / s

Step-by-step explanation:

The simple harmonic motion of a spring and a mass is described by the equation

x = A cos (wt + Фfi)

where A is the amplitude of movement in this case 0.25 m, w the angular velocity and fi the initial phase

the angular velocity is

w = √ k / m

We can use Hooke's law to find the constant ka

F = k x

where the force is 300N and the spring stretch is 0.25m

k = F / x

k = 300 / 0.25

k = 1200 N / m

To find the phase angle di, let's use the system speed

va = dx /dt

va = A w sin (wt + Ф)

they tell us that the spring comes out of rest at time zero

Vd = Aw sin Ф

the only way this term is zero is that the angle Ф = 0

substitutions in the first equations

x = A cos wt

with

w = √RA (1200 / 0.5)

w = 48.99 rad / sec

we substitute in the first equations

x = 0.25 cos (48.99 t)

speed is

v = 0.25 48.99 without 48.99i

ask the speed for x = 0.15 m

we start by calculating the time it takes to get to this point

x = A cos wt

t = 1 / w cos-1 x / A

we look for the time

t = 1 / 48.99 cos-1 (0.15 / 0.25)

t = 0.0189 s

this is the first time it takes to get to the requested point

now we can calculate the speed

v = Aw sin (wt)

v = 0.25 48.99 sin (48.99 0.0189)

v = 11.34 m / s

User CostelloNicho
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