Problem

The prongs of a tuning fork each vibrate with an …

01:50

Problem 44 Medium Difficulty

A $5.0 \mathrm{kg}$ block hangs from a spring with spring constant $2000 \mathrm{N} / \mathrm{m} .$ The block is pulled down $5.0 \mathrm{cm}$ from the equilibrium position and given an initial velocity of $1.0 \mathrm{m} / \mathrm{s}$ back toward equilibrium. What are the (a) frequency, (b) amplitude, and (c) total mechanical energy of the motion?

Answer

b. $7.1 \mathrm{cm}$
c. $5.0 \mathrm{J}$

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Physics 101 Mechanics

Physics for Scientist and Engineers A Strategic Approach

Chapter 14

Oscillations

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Christina K.

Rutgers, The State University of New Jersey

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Video Transcript

here for part A. We know that the frequency F is equaling 1/2 pi multiplied by the square root of the spring Constant K divided by the Mass M. This is equally 1/2 pi multiplied by the square root of 2000 Newtons per meter. This will be divided by 5.0 kilograms and we find that then the frequency F is approximately 3.18 hurts For part B. We can find the amplitude from energy conservation. This would be equal in the square root of the initial position squared, plus the initial velocity divided by omega quantity squared. And so this is equaling square root of 0.500 meat. Rather, my family's 0.50 meters quantity squared plus are 1.0 meters per second, divided by two pi multiplied by 3.183 hurts again. We're round at the very end and this terms where stand the square root and we find that then the amplitude A is equaling 2.0 707 meters. This is essentially 7.7 centimeters, and then for part c, we can see that the total mechanical energy would be equaling 2 1/2 times this spring, constant multiplied by the amplitude squared. This would be 1/2 multiplied by 2000 Newtons per meter multiplied by point 0707 meters. Quantity squared. And we find that then the total mechanical energy is equaling, then approximately 5.0 jewels. That is the end of the solution. Thank you for watching.

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