Question

PrintCalculator Periodic Table Question 27 of 40 A physics lab is demonstrating the principles of simple harmonic motion (SHM) by using a spring affixed Mapd a horizontal support. The student is asked to find the spring constant, k. After suspending a mass of 255.0 g from the spring, the student notices the spring is displaced 37.5 cm from its previous equilibrium. With this information, calculate the spring constant. Number N m When the spring, with the attached 255.0-g mass, is displaced from its new equilibrium position, it undergoes SHM. Calculate the period of oscillation, T, neglecting the mass of the spring itself. Number

Answer 1

1) mass attached m = 255 gram;

weight of the mass (w) = m*g = 0.255 kg * 9.8 m/s2 = 2.499 N;

force acting on the spring (F) = K*X; where X is the dispacement of the spring from equilibrium position;

= weight of the mass (w);

2.499 N = K*0.375 m;

Stiffness of the spring, K = 6.664 N/m.

2) time period T = 2*pi* squareroot(m/ K);

T = 2*pi* squareroot(0.255/ 6.664);

T = 1.23 sec.

3) At equilibrium displacement is zero (x = 0);

now spring is displaced by x = 28.1 cm from the equilibrium position;

total energy of the system (E) = KE + PE

At x = 28.1 cm:

KE will be zero.

so velocity V = 0 m/s;

potential energy stored in the spring will be U = 0.5*K*x²;

U = 0.5* 6.664* (0.281)²;

U = 0.263 J.

total energy = 0 + 0.263 = 0.263 J.

At x = 20.6 cm: (at this instant there will be both type of energy KE and PE)

potential energy U = 0.5*K*x²;

U = 0.5* (6.664)* (0.206²);

U = 0.141 J.

now KE = TE - PE = 0.263 - 0.141;

KE = 0.1216 J;

0.1216 = 0.5* m* v²;

v = 0.976 m/s.