Question

(a) a spring stretches by 0.015 m when a 1.75 kg object is suspended from its end. what is its spring constant? (b) how much mass should be attached to the spring so that its frequency of vibration is f

Answer 1

(a) Total force along y axis,

∑F =ma=0

kx−mg=0 ⇒k= $\frac{mg}{x}$

The spring constant is, k is

k=(1.5)×(9.8)/(0.020)=14.7/0.020 =735N/m

(b) Frequency f =  $\frac{1}{2} \pi$ $\sqrt{\frac{k}{m} }$  ⇒ 4$\pi ^{2}$$f^{2}$ = $\frac{k}{m}$

$\frac{k}{4\pi ^{2}f^{2} }$= m ⇒ m = (735)4$\pi ^{2}$(3.0)(0.3)

=(735)×4×(3.75)×(3.75)×(9.0)

=735×4×9.85×9.0=7354×9.85×9.0

=735354.6 N

=2.07kg

The spring constant is the force required to stretch or compress the spring divided by the distance the spring is lengthened or shortened. It is used to determine the stability or instability of a spring and thus the system for which it is intended. It is denoted by k.

Therefore the spring constant is 735 N/m and the mass should be 2.07kg.

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