Question

A 130g oscillator has a speed of 124cm/s when its displacement is 2.10cm and 45.5cm/s when its displacement is 5.70cm . What is the oscillator's maximum speed?

Answer 1

We know that ET never changes:
ET= EK + EP = 1/2 mv^2 + 1/2 kx^2

And we are given the speeds, v1 and v2, at two positions, x1 and x2, so:
ET1 = ET2
1/2 m(v1)^2 + 1/2 k(x1)^2 = 1/2 m(v2)^2 + 1/2 k(x2)^2

Every variable is given except k. so you can solve for it:
1/2 m(v1)^2 - 1/2 m(v2)^2 = 1/2 k(x2)^2 - 1/2 k(x1)^2
1/2 m(v1)^2 - 1/2 m(v2)^2 = 1/2 k(x2)^2 - (x1)^2)
k = m((v1)^2 -(v2)^2) / ((x2)^2 - (x1)^2

Answer 2

Mass m = 130 g

              = 0.130 kg

Speed at x=2.10cm is v = 124cm / s

We know v = ? ?[A ² - x ² ]

          124 = ??[A ² - 2.1 ² ] -----------( 1)

Also given speed at x '= 5.7 cm is v ' =45.4 cm / s

i.e., v ' = ??[A ² - x' ² ]

     45.5 = ? ?[A ² - 5.7 ² ]       --------( 2)

eq( 1) / eq( 2 ) ==>

                                  2.73 = [A ² - 2.1 ² ]/[A ² - 5.7 ² ]

                            A = 48.72 cm

Substitue this value in eq( 1) you get , 124= ? x 48.68

                                                               ? = 2.55rad / s

? Maximum velocity = A ?

                                  = 124.11 cm/ s