Question

A 151 g oscillator has a speed of 130 cm/s when its displacement is 2.00 cm and 75.5 cm/s when its displacement is 5.10 cm . What is the oscillator's maximum speed?

Answer 1

Let A is the Amplitude of motion.

Apply conservation of energy

(1/2)*k*A^2 = (1/2)*m*v^2 + (1/2)*k*x^2

(1/2)*m*v^2 = (1/2)*k*A^2 - (1/2)*k*x^2

v^2 = (k/m)*A^2 - (k/m)*x^2

v^2 = w^2*A^2 - w^2*x^2

v = w*sqrt(A^2 - x^2)

in the first case,

v1 = w*sqrt(A^2 - x1^2)

130 = w*sqrt(A^2 - 2^2)   ---(1)

in the second case

75.5 = w*sqrt(A^2 - 5.1^2) ---(2)

take equation(1)/equation(2)

130/75.5 = sqrt(A^2-2^2)/sqrt(A^2 - 5.1^2)

(130/75.5)^2 = (A^2-2^2)/(A^2 - 5.1^2)

2.9647 = (A^2 - 4)/(A^2 - 26.01)

(A^2 - 26.01)*2.9647 = A^2 - 4

2.9647*A^2 - 77.11 = A^2 - 4

1.9647*A^2 = 77.11 - 4

A^2 = 73.11/1.9647

A = sqrt(73.11/1.9647)

= 6.1 cm

let k is spring constant.

in the first case

0.5*k*A^2 = 0.5*k*x1^2 + 0.5*m*v1^2

0.5*k*(A^2 - x1^2) = 0.5*m*v1^2

k = m*v1^2/(A^2 - x1^2)

= 0.151*1.3^2/(0.061^2 - 0.02^2)

= 76.84 N/m

angular frequency of motion, w = sqrt(k/m)

= sqrt(76.84/0.151)

= 22.56 rad/s

Vmax = A*w]

= 6.1*22.56

= 137.6 cm/s or 1.376 m/s <<<<<<<<-----Answer