When an unknown weight W was suspended from a spring with an unknown force const

Question

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 9.67 cm because of
the weight W .
Then the weight W was pulled further down
to a position 13 cm (3.33 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.

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 9.67 cm because of the weight W . Then the weight W was pulled further down to a position 13 cm (3.33 cm below its equilibrium position) and released, which caused an oscillation in the spring.

Explanation / Answer

k = w / 0.0967 N/m
k = mg / 0.0967

where m = mass

m/k = 0.0967 / 9.8 = 0.00986

period of SHM
= 2 pi sq root [ m/k]

= 2 * pi * sq rt (0.00986)

= 0.624 s

Frequency = 1/ 0.624 = 1.6 Hz

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