A 200 g block attached to a horizontal spring is oscillatingwith an amplitude of

Question

A 200 g block attached to a horizontal spring is oscillatingwith an amplitude of 2.0 cm and a frequency of 2.0 Hz. Just as itpasses through the equilibrium point, moving to the right, a sharpblow directed to the left exerts a 20 N force for 1.0 ms. a) What is the new frequency? b) What is the new amplitude? A 200 g block attached to a horizontal spring is oscillatingwith an amplitude of 2.0 cm and a frequency of 2.0 Hz. Just as itpasses through the equilibrium point, moving to the right, a sharpblow directed to the left exerts a 20 N force for 1.0 ms. a) What is the new frequency? b) What is the new amplitude?

Explanation / Answer

a)the new frequency f = (1/t) ---------------(1) t = 1.0 ms = 1.0 * 10^-3 s b)the speed of the block is (1/2)m * v^2 = F * A or v = (2F * A/m)^(1/2) F = 20 N,A = 2.0 cm = 2.0 * 10^-2 m and m = 200 g = 200 *10^-3 kg We know from the relation vmax= A * w or A = (vmax/w) vmax= v and w = 2f,where the value of f isobtained from equation (1)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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