contestada

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 21 cm because of the

weight W .

Then the weight W was pulled further down

to a position 31 cm (10 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/s2.

Answer in units of Hz

Respuesta :

mirianmoses

Answer:

f= 1/2pi*sqrt(g/static elongation)= 1.08 Hz

Explanation:

explanation at equilibrium mg= k*static elongation k= mg/static elongation frequency= sqrt(k/m)/2pi= sqrt(g/static elongation)/2pi

OR

k = w / 0.21N/m

= mg / 0.21

where m = mass

m/k = 0.21 / 9.8 = 0.0214

period of SHM

= 2 pi sq root [ m/k]

= 2 * pi * sq rt 0.0214

= 0.919 s

Frequency = 1/ 0.919 = 1.08 Hz

Otras preguntas