Question

Consider the spring-mass system described by the equation my''+By+ky=0, where y denotes the distance from equilibrium. Suppose that mass is 3kg and that there is no friction. The spring is stretched by 0.1 m and then struck with a hammer so that y(0)=-0.1 and y'(0)=1, as a result, the mass begins to oscillate at the rate of 4 radians per second. a) What is the spring constant? Justify your answer by computing the homogeneous solution. b) What is the amplitude of the oscillation?

Answer 1

(a) The spring constant can be calculated from the formula for the angular frequency. The angular frequency of oscillation is given by,

$\omega = \sqrt{\frac{k}{m}} \Rightarrow 4 = \sqrt{\frac{k}{3}} \Rightarrow k = 48 \, N/m$

(b) The amplitude will be maximum when the spring will be stretched at the max. The total energy of the system will be always conserved. The initial energy of the system is, (substituting initial position and velocity)

$E_i = \frac{1}{2}k(-0.1)^2 + \frac{1}{2}m(1)^2$

Substituting values we get,

2 Eiー2(48)(-0.1)2+-(3)(1)-1.74 J

At the amplitude position, let us say that the stretch is A. At that point the particle will not have any kinetic energy. Hence the energy at the state will be,

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

By Law of conservation of energy, we can write,

$\Rightarrow \frac{1}{2}(48)A^2 = 1.74 \Rightarrow A = 0.27 \, m$

Hence the amplitude will be 0.27 m.