A spring balance has a scale that reads from 0 to 50 kg.The length of the scale is 20 cm.A block of mass #m# is suspended from this balance,when displaced and released,it oscillates with a period 0.5 s. What is the value of #m#?[#g=10 "m" s^(-2)#]

1 Answer
Ananda Dasgupta
Mar 28, 2018

#~~15.8" Kg"#

Explanation:

I am guessing that the wording of the question means that the spring stretches from 0 to 20 cm as the weight increases from 0N to 500 N. This means that the spring constant is

#k = (500" N")/(0.2" m") = 2500" Nm"^-1#

The time period of oscillations of a mass #m# attached to a spring with spring constant #k# is given by

#T = 2pisqrt(m/k)#

Thus

#m=k(T/(2pi))^2 ~~2500" Nm"^-1 ((0.5" s")/(2times 3.14))^2 =15.8" Kg"#

Related questions
Impact of this question
1429 views around the world
You can reuse this answer
Creative Commons License