Question

A 0.275 kg block rests on a frictionless level surface and is attached to a 20 pts. horizontally aligned spring with a spring constant of 650. N/m. The block is initially displaced 4.00 cm from the equilibrium point and then released to set up a simple harmonic motion. Find: a) the total mechanical energy of the vibration; b) the speed of the block when it passes through the equilibrium point; c) the period and frequency of the vibration

Answer 1

a) E = 0.52 J

b) v = 1.94 m/s

c) T = 0.129 s

f = 7.73 s^-1

Step-by-step explanation:

a) To find the total mechanical energy you use the following formula:

`E=(1)/(2)kA^2` (1)

k: spring constant = 650N/m

A: amplitude of the oscillation = 4.00cm = 0.04m

you replace the values of the parameters in (1) to calculate E:

`E=(1)/(2)(650N/m)(0.04m)^2=0.52J`

the total mechanical energy is 0.52J

b) To find the speed in the position of equilibrium you take into account that at that point all mechanical energy is kinetic energy, then, you se the following equation:

`E=K=(1)/(2)mv^2`

m: mass of the block = 0.275 kg

`K=0.52J=(1)/(2)mv^2\\\\v=\sqrt{(2(0.52J))/(0.275kg)}=1.94(m)/(s)`

c) The period and frequency are given by:

`T=2\pi\sqrt{(m)/(k)}=2\pi\sqrt{(0.275kg)/(650N/m)}=0.129s\\\\f=(1)/(T)=(1)/(0.129s)=7.737s^(-1)`