Answer:
a) [tex]k = 200.016\,\frac{N}{m}[/tex], b) [tex]m = 1.389\,kg[/tex], c) [tex]f = 0.524\,hz[/tex]
Explanation:
a) The maximum speed of the oscillating block-spring system is:
[tex]v_{max} = \omega \cdot A[/tex]
The angular frequency is:
[tex]\omega = \frac{v_{max}}{A}[/tex]
[tex]\omega = \frac{1.2\,\frac{m}{s} }{0.1\,m}[/tex]
[tex]\omega = 12\,\frac{rad}{s}[/tex]
The mass of the system is:
[tex]E = \frac{1}{2}\cdot m\cdot v_{max}^{2}[/tex]
[tex]m = \frac{2\cdot E}{v_{max}^{2}}[/tex]
[tex]m = \frac{2\cdot (1\,J)}{(1.2\,\frac{m}{s} )^{2}}[/tex]
[tex]m = 1.389\,kg[/tex]
The spring constant is:
[tex]\omega = \sqrt{\frac{k}{m} }[/tex]
[tex]k = \omega^{2}\cdot m[/tex]
[tex]k = (12\,\frac{rad}{s} )^{2}\cdot (1.389\,kg)[/tex]
[tex]k = 200.016\,\frac{N}{m}[/tex]
b) The mass is:
[tex]m = 1.389\,kg[/tex]
c) The frequency of oscillation is:
[tex]\omega = 2\pi\cdot f[/tex]
[tex]f = \frac{2\pi}{\omega}[/tex]
[tex]f = \frac{2\pi}{12\,\frac{rad}{s} }[/tex]
[tex]f = 0.524\,hz[/tex]
Answer:
a) F = 20 N
b) m = 1.39 kg
c) f = 1.909 Hz
Explanation:
Given
E = 1 J
A = 0.1 m
vmax = 1.2 m/s
a) F = ?
b) m = ?
c) f = ?
Solution
a) We apply the equation
E = 0.5*k*A²
then
k = 2*E/A²
k = 2*1 J/(0.1 m)²
k = 200 N/m
then we use the equation
F = kA
F = (200 N/m)(0.1 m)
F = 20 N
b) We use the formula
E = K + U
if U = 0 J
then
E = K = 0.5*m*v²
⇒ m = 2*K/v²
m = 2*1 J/(1.2 m/s)²
m = 1.39 kg
c) we apply the equation
f = (1/2π)√(k/m)
then
f = (1/2π)√(200 N/m/1.39 kg)
f = 1.909 Hz
