Next: About this document ...
Up: Vibrations and Waves
Previous: The Simple Pendulum
A 0.5 kg mass is hung on a vertical massless spring. The new
equilibrium position
of the spring is found to be 3 cm below the equilibrium position
of the
spring without the mass. a) What is the spring constant, k ? b)
Show that the
mass and spring system oscillates with simple harmonic motion about
the new
equilibrium position.
Solution:
F = 0 = kx_{ eq} - mg | (18) |
k | = | ||
= | = 163 N/m . |
F_{s} = k(x_{ eq} - s)
The total force upward is thereforeF_{ tot} = F_{s} - mg = kx_{ eq} - ks - mg
Since the equilibrium position obeys Eq.(13.18), the equation for the acceleration of the mass reduces to:F_{ tot} = - ks | (19) |
Consider a SHO with m = 0.5 kg, k = 10 N/m and amplitude
A = 3
cm. a) What is the total energy of the oscillator? b) What is its
maximum
speed? c) What is the speed when x = 2 cm? d) What are the kinetic
and
potential energies when x = 2 cm?
Solution:
E_{ total} = kA^{ 2} = (10)(0.03) = 0.0045 J
v_{ max} = = = 0.134 m/s
v | = | ||
= | = 0.10 m/s |
PE_{s} | = | kx^{ 2} = (10)((0.02)^{2}) = 0.002 J | |
KE | = | mv^{ 2} = (0.5)((0.1)^{2}) = 0.0025 J |
If the oscillator in the above problem is released from rest
at x = A
when the clock is set to t = 0 seconds, a) determine the position
and
velocity of the oscillator at t = 2 s. b) At what time does the
oscillator get to x = - 1 cm?
Solution:
x(t) | = | Acos | |
= | (0.03)cos - 0.027 m |
v(t) | = | Asin | |
v(2) | = | (0.03)sin 0.062 m/s |
x(t) | = | Acos | |
t | = | ||
= | 0.43 s |
If the mass in the above problem experiences a constant
frictional
force of 0.1 N and is released from rest at x = A , a) what is
the velocity
of the mass when it first passes through the equilibrium position
at x = 0 . b) At what
value of x does the velocity of the mass first go to zero after
it has been
released?
Solution:
W_{ ext} | = | KE + PE_{s} + PE_{ grav} | |
- F_{ fr}d | = | 0 | |
v_{f} | = | ||
= | = 0.077 m/s |
W_{ ext} | = | KE + PE_{s} + PE_{ grav} | |
- F_{ fr}d | = | 0 + 0 | |
- F_{ fr}(A - x_{0}) | = | ||
0 | = | kx_{0}^{2} - 2F_{ fric}x_{0} - kA^{ 2} + 2F_{ fric} | |
= | (10)x_{0}^{2} - 2(0.1)x_{0} + | ||
= | (10)x_{0}^{2} - 0.2x_{0} - 0.003 |
x_{0} | = | m | |
= | - 0.01 m or 0.03 m | (20) |
A simple pendulum is used in a
physics laboratory experiment to obtain an experimental value for
the
gravitational acceleration, g . A student measures the length of
the
pendulum to be 0.510 meters, displaces it 10^{ o } from the
equilibrium
position, and releases it. Using a stopwatch, the student
determines that
the period of the pendulum is 1.44 s. Determine the experimental
value of
the gravitational acceleration.
Solution:
Since the angle through which the pendulum is initially displaced
is small,
Eq.(13.17) in the Lecture Notes is valid for the period T :
T | = | 2 | |
g | = | ||
= | = 9.71 m/s ^{2} |
www-admin@theory.uwinnipeg.ca