A block of wood of mass 53 g floats in a swimming pool, oscillating up and down

Question

A block of wood of mass 53 g floats in a swimming pool, oscillating up and down in simple harmonic motion with a frequency of 4 Hz. (a) What is the value of the effective spring constant of the water? (b) A partially filled water bottle of almost the same size and shape as the block of wood but with mass 275 g is placed on the water's surface. At what frequency will the bottle bob up and down?

Explanation / Answer

We generally do not answer the questions of those who are having low rating. You are having a rating of "92%". Please rate all answers and improve your rating. Even if you find the answers are incorrect, first rate Lifesaver or helpful to the best answer and then rate other answers as not helpful. But do rate them for the good of both of us. Thanks!! Coming to the problem m = 53 g = (53/1000) Kg f = 4 Hz = 1/T =>T = 1/4 seconds w = 2pi/T = ((2*pi())/(1/4)) a)w = sqrt(k/m) =>k = m*w^2 = (53/1000)*(((2*pi())/(1/4))^2) = 33.478 N/m b)w = sqrt(k/m) But Now m = 275 g = (275/1000) =>w = sqrt(k/(275/1000)) = 11.033 =>2pif = 11.033 =>f = 1.756 Hz

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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