Question

A 0.50-kg box is attached to an ideal spring of force constant (spring constant) 20 N/m on a horizontal, frictionless floor. The box oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. (a) What is the amplitude of vibration

Answer 1

Answer:

$A = 0.24 m$

Explanation:

As we know that for spring block system the angular frequency is given as

$\omega = \sqrt{\frac{k}{m}}$

now we know that

$k = 20 N/m$

also we know that

$m = 0.50 kg$

so we have

$\omega = \sqrt{\frac{20}{0.5}}$

$\omega = 6.32 rad/s$

now we know that speed of an SHM at its equilibrium position is given as

$v = A\omega$

now plug in all values in it

$1.5 = A(6.32)$

$A = 0.24 m$