Question

A mass of 0.30 kg on the end of a spring oscillates with a period of 0.45 s and an amplitude of 0.15 m . A) Find the velocity when it passes the equilibrium point. B) Find the total energy of the system. C) Find the spring constant. D) Find the maximum acceleration of the mass.

Answer 1

let position at any time be given as

x(t)=A*sin(w*t)

where A=amplitude=0.15 m

w=angular frequency=2*pi/period=2*pi/0.45=13.963 rad/sec

hence x(t)=0.15*sin(13.963*t) m

veloicty=dx/dt=0.15*13.963*cos(13.963*t)=2.0944*cos(13.963*t) m/s

as angular frequency=sqrt(spring constant/mass)

==>13.963=sqrt(spring constant/0.3)

==>spring constant=13.963^2*0.3=58.49 N/m

a)

when the mass passes through equilibrium, x(t)=0

==>w*t=0

at that time, speed=2.0944*cos(0)=2.0944 m/s

part b:

as there is no friction, total energy of the system will be conserved.

at the highest displacement point, speed of the mass is 0.

then total energy of the system=potential energy of the spring=0.5*spring cosntant*amplitude^2

=0.5*58.49*0.15^2=0.658 J

part c:

spring constant is 58.49 N/m

part d:

acceleration of the mass=dv/dt=-2.0944*13.963*sin(13.963*t)

as maximum value of sine function is 1,maximum acceleration=2.0944*13.963=29.244 m/s^2