A mass of 0.56 kg is attached to a spring and set into oscillation on a horizont

Question

A mass of 0.56 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic motion of the mass is described by

x(t) = (0.24 m)cos[(6 rad/s)t].

Determine the following.

(a) amplitude of oscillation for the oscillating mass
m

(b) force constant for the spring
N/m

(c) position of the mass after it has been oscillating for one half a period
m

(d) position of the mass one-third of a period after it has been released
m

(e) time it takes the mass to get to the position

x = ?0.10 m

after it has been released
s

Explanation / Answer

A) amplitude is A = 0.24m...

B) w = sqrt(k/m)

spring constant k = m*w^2 = 0.56*36 = 20.16 N/m.....

C) x = A*cos(wt)...
when t = T/2...
and w = 2*pi/T

x = A*cos(w*T/2) = A*cos(pi) = -A = -0.24m

D) when t = T/3...
x= A*cos(wt) = A*cos(2*pi/3) =-0.5*A = -0.12m

e) when x = -0.1m...

-0.1 = 0.24*cos(6t)...

cos(6t) = -0.416...

6t = acos(-0.416) = 1.99...
t = 1.99/6 = 0.333 sec

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.