Question

A simple pendulum has a length L and a mass m. At its highest point, the pendulum mass is 0.25L above its lowest point (see figure below). What is the speed of the mass when it is at its lowest point? Express your answer in terms of m, L, and g. v = The position of a mass-on-a-spring oscillator is given by y = A sin(20t), where the value of t is in seconds and A = 0.44 m. What is the maximum kinetic energy of an oscillator of mass 2.2 kg? Suppose the amplitude is increased so that the maximum kinetic energy is doubled. What is the new value of A? m

Answer 1

here,

length is l and mass is m

height difference , h = 0.25 * l

let the speed of mass at the lowest point be v

using conservation of energy

0.5 * m * v^2 = m * g * ( h)

0.5 * v^2 = 9.8 * 0.25 * l

v = 2.21 sqrt(l) m/s

----------------------------

y = A * sin(20 * t)

A = 0.44 m

(a)

m = 2.2 kg

maximum speed , vm = A * w

vm = 0.44 * 20 = 8.8 m/s

the maximum kinetic energy of the mass is , KE = 0.5 * m * vm^2

KE = 0.5 * 2.2 * 8.8^2

KE = 85.18 J

(b)

let the new amplitude be A'

maximum speed , vm' = A' * w

vm' = A' * 20

2 * KE = 0.5 * m * vm'^2

2 * 85.18 = 0.5 * 2.2 * (A'*20)^2

A' = 0.62 m

the new amplitude is 0.62 m