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Get Full Access to University Physics - 13 Edition - Chapter 14 - Problem 18e
Get Full Access to University Physics - 13 Edition - Chapter 14 - Problem 18e

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# A 0.500-kg mass on a spring has velocity as a function of

ISBN: 9780321675460 31

## Solution for problem 18E Chapter 14

University Physics | 13th Edition

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Problem 18E

A 0.500-kg mass on a spring has velocity as a function of time given by vx(t) = -(3.60 cm/s) sin[( 4.71 rad/s)t – (?/2)]. What are (a) the period; (b) the amplitude; (c) the maximum acceleration of the mass; (d) the force constant of the spring?

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Solution 18E The expression for velocity of an object executing SHM is given by, v(x) = Asin(t + ), symbols have usual meaning. Comparing this equation with the given equation, vx(t) = -(3.60 cm/s) sin[( 4.71 rad/s)t – (/2)] Comparing these two expressions, we get A = 3.60 cm/s , = 4.71 rad/s. (a) o, 2/T = 4.71 rad/s T = 1.33 s The period is 1.33 s. (b) Amplitude, A = 3.60cm = 0.76 cm 4.71 Therefore, the amplitude is 0.76 cm. (c) Acceleration will be maximum when the displacement is maximum, or the at the amplitude. Maximum acceleration, a = A2 max 2 a max = .A = 4.71 rad/s × 3.60 cm/s = 16.956 cm/s 2 The maximum acceleration is 16.956 cm/s . (d) Mass of the object m = 0.500 kg 2 2 2 2 Therefore, spring constant k = m = (4.71) rad /s × 0.500 kg = 11.09 N/m Therefore, the spring constant is 11.09 N/m.

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