A block of mass 650 g on the end of spring undergoes simple

A block of mass 6.50 g on the end of spring undergoes simple harmonic motion with a frequency of 9.00 Hz. What is the spring constant of the spring? 2.079 x101 N/m You are correct. Previous Tries Your receipt no. s 162-6163 (2 If the motion of the mass has an initial amplitude of 8.50 cm what is its maximum speed?. 4.807x 102 cm/s You are correct. Previous Tries Your receipt no s 162-6930 (2 The amplitude decreases to 2.529 cm in 1.44 s, what is the damping constant for the system? 01145

Solution

Here ,

m = 6.5 gm = 0.0065 Kg

frequency , f = 9 Hz

let the spring constant is k

as f = 1/(2pi) * sqrt(k/m)

9 = 1/(2pi) * sqrt(k/.0065)

k = 20.78 N/m

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A = 8.5 cm = 0.085 m

let the maximum speed is v

0.5 * m * v^2 = 0.5 * k * A^2

0.0065 * v^2 = 20.78 * 0.085^2

v = 4.807 m/s

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as A = Ao * e^(-bt)

2.529 = 8.5 * e^(-b * 1.440)

solving for b

b = 0.841 1/s

the damping constant is 0.841 1/s