Question

A simple harmonic oscillator consists of a block of mass 2.00 kg attached to a spring of spring constant 100 N/m. When t = 1.00 s, the position and velocity of the block are x = 0.129 m and v = 3.415 m/s respectively.

a) What is the amplitude of oscillations?

b) What were the position and velocity of the mass at time t = 0?

Answer 1

given that

m = 2 kg

k = 100 N/m

t = 1 s

x = 0.129 m

v = 3.415 m/s

from the equation of simple harmonic motion

x = A*sin(w*t + phi)      ......eq1

A = x/sin(w*t + phi)        .......eq2

v = w*A*cos(w*t + phi)     ........eq3

where , phi = phase shift

we know that

w = sqrt(k/m) =sqrt (100 / 2) = 7.07 rad/s

put the value of A from eq2 into eq3

v = w * x/sin(w*t + phi) * cos(w*t + phi)

from trignometry

cos(theta) / sin(theta) = cot(theta) = 1/tan(theta)

v = w * x / tan(w*t + phi)

tan(w*t + phi) = w*x/v

w*t + phi = tan^-(w*x/v)

phi = tan^-(w*x/v) - w*t

phi = tan-(7.07 *0.129/3.415) - 7.07*1

phi = -6.82 rad

we know that, 2*pi rad = 6.28

but the angle is more than 6.28 rad so actual phi = -6.81 + 6.28 = -0.53 rad

from eq2

A = x/sin(w*t + phi)

A = 0.129 / sin(7.07*1 - 0.53)

A = 0.50 m

(b)

x = A*sin(w*t + phi)

at t=o

x = 0.50*sin(0-0.53)

x = -0.252 m

(c)

at t = 0

v = w * A*cos(w*t + phi)

v = 7.07*0.50 *cos(0-0.53)

v = 3.05 m/s