Question

A 120 g block attached to a spring with spring constant 3.0 N/m oscillates horizontally on a frictionless table. Its velocity is 21 cm/s when x0 = -4.1 cm

Answer 1

Answer:

0.059 m

Explanation:

$0.5k A^2 = 0.5 (k x^{2} + m v^{2})$ where k is the spring constant, A is the amplitude, x is the extension of spring, m is the mass of the spring and v is the velocity in m/s. Making a the subject then

$A = \sqrt{(x^{2} + (\frac {mv^{2}}{k})}$

By substituting the given values then where m is 0.12 Kg, x= 0.0041 m, v is 0.21 m/s and K is 3 N/m then

$A = \sqrt{(0.0041^{2} + (\frac {0.12\times 0.21^{2}}{3})}=0.058694122m\approx 0.059 m$