When an unknown weight W was suspended from a spring with an unknown force constant k it reached its equilibrium position and the spring was stretched by 58.6 cm because of the weight W . Then the weight W was pulled further down to a position 71 cm (12.4 cm below its equilibrium position) and released, which caused an oscillation in the spring. Calculate the cyclic frequency of the resulting motion. The acceleration due to gravity is 9.8 m/s 2 . Answer in units of Hz.

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explanation:

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s