An object of mass 1.5 kg is supported by a vertical lossless spring of force con

Question

An object of mass 1.5 kg is supported by a vertical lossless spring of force constant 1800 Nm^-1. When pulled down 2.5 cm from equilibrium and released from rest the object oscillates with simple harmonic motion such that its displacement from equilibrium, x, is given by x = A cos(omega t + delta), where omega = squareroot k/m Define the terms A, omega, and delta Determine the amount the spring is stretched from its natural length when the object is in equilibrium. the values of A, omega, and delta. the speed and acceleration of the object when t = 0.5 s. This mass-spring system is then set out horizontally on a frictionless surface. It is then set into oscillation with the same amplitude as the vertical case. Sketch the potential energy curve for the system. Determine the following values and show them on your curve the total energy, E, the kinetic energy, K, at t = 0.25 s the potential energy, U, at t = 0.25 s Find the first three times at which the kinetic energy, K, equals the total energy, E.

Explanation / Answer

a) here,

A is the amplitude

w is the angular frequency

delta is the phase constant

b)

i) let the spring is stretched by x

for equilibrium

k * x = m * g

1800 * x = 1.5 * 9.8

x = 8.17 *10^-3 m

the spring is stretched by 8.17 *10^-3 m

ii)

the values are

A = 2.5 cm = 0.025 m

w = sqrt(k/m)

w = sqrt(1800/1.5)

w = 34.64 rad/s

Now,as the mass is initially at the extreme position

delta (phase constant) = pi

iii)
at t = 0.5 s

v = -A * w * sin(w * t)

v =- 0.025 * 34.64 * sin(34.64 * 0.5)

v = 0.865 m/s

the velocity at t = 0.5 s is 0.865 m/s

for the acceleration

a = A * w^2 * cos(w * t)

a = - 0.025 * 34.64^2 * cos(34.64 * 0.5)

a = -1.24 m/s^2

the acceleration at t = 0.5 s is -1.24 m/s^2

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