Question

A 5.30kg block hangs from a spring with a spring constant 1700 N/m. The block is pulled down 4.50cm from the equilibrium position and given an initial velocity of 1.70 m/s back toward equilibrium. a) what is the frequency of the motion (units: Hz) b) what is the amplitude (units: m) c) what is the total mechanical energy of the motion (units: J)

Answer 1

To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)

PART A) By definition the frequency in a spring is given by the equation

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

Where,

m = mass

k = spring constant

Our values are,

k=1700N/m

m=5.3 kg

Replacing,

$f = \frac{1}{2\pi} \sqrt{\frac{1700}{5.3}}$

$f=2.85 Hz$

PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

$\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} kY^2$

Where,

k = Spring constant

m = mass

y = Vertical compression

v = Velocity

This expression is equivalent to,

$kA^2 =mV^2 +ky^2$

Our values are given as,

k=1700 N/m

V=1.70 m/s

y=0.045m

m=5.3 kg

Replacing we have,

$1700*A^2=5.3*1.7^2 +1700*(0.045)^2$

Solving for A,

$A^2 = \frac{5.3*1.7^2 +1700*(0.045)^2}{1700}$

$A ^2 = 0.011035$

$A=0.105 m \approx 10.5 cm$

PART C) Finally, the total mechanical energy is given by the equation

$E = \frac{1}{2}kA^2$

$E=\frac{1}{2}1700*(0.105)^2$

$E= 9.3712 J$