You pull the 0.5-kg hanging mass 10 cm down from its equilibrium position. When

Question

You pull the 0.5-kg hanging mass 10 cm down from its equilibrium position. When you release it, it starts oscillating to & fro.
1)Calculate the characteristic frequency of its natural oscillations. (round the answer to the whole number). Stiffness constant of the spring, k = 20 N/m.
Hint: use the formula for natural frequency of simple harmonic oscillations, f. P.S. This frequency is called natural because if you disturb the mass from equilibrium and let it move by itself, its oscillations will be NATURAL.


2) Calculate the period of its oscillations (round the answer to the whole number). Stiffness constant of the spring, k = 20 N/m. Hint: period of simple harmonic oscillations, T = 1/f, where f is frequency.

Explanation / Answer

The hanging mass m = 0.5kg the spring constant k = 20N/m (1) The characteristic frequency                        = k/m                            = (20N/m) /(0.5kg)                            = 6.3 rad/s therefore the frequency                 f = /2                   = 6.3rad/s / 2                   = 1Hz (2) The period of its oscillations               T = 1/f = 1/1 = 1sec

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.