Question

Un resorte se monta horizontalmente con su extremo izquierdo fijo. Conectando una balanza de

Answer 1

Translation of important question part (Google translation used)

Force of 6N causes a displacement of 0.03m. remove the balance and connect a body 0.5 kg to the end, pull it to move it 0.02 m, release it and see how it oscillates.

(a)Determine the spring constant.

(b)Calculate angular velocity, frequency, and oscillation period

Answer:

(a)K=200 N/m

(b) w= 20 rad/s f=3.2 Hz T=0.3125 s

Explanation:

(a)

From Hooke's law, we deduce that F=kx where F is applied force, k is spring constant and x is extension of the spring. Making k the subject of the formula then k=F/x and substituting F with 6 N and x with 0.03 m then k=6/0.03=200 N/m

(b)

Angular velocity, w is given by $w= \sqrt{\frac {k}{m}}$ where m is the mass and k is spring constant calculated in part a above. Substituting mass with 0.5 kg and k with 200 N/m then

$w= \sqrt{\frac {200}{0.5}}=20 rad/s$

We know that frequency, f is given by $f=\frac {w}{2\pi}$ and substituting 20 rad/s for w then

$f=\frac {20}{2\pi}=3.1830988618379067153776752674502872406891\approx 3.2 Hz$

Finally, oscillation period, T is usually the reciprocal of frequency hence T=1/f and substituting f with 3.2 Hz then T=1/3.2=0.3125 s