PHYSICS 1 Worksheet Ch 15 Review ***PLEASE ANSWER AS MANY QUESTIONS AS POSSIBLE!
ID: 1620015 • Letter: P
Question
PHYSICS 1
Worksheet Ch 15 Review
***PLEASE ANSWER AS MANY QUESTIONS AS POSSIBLE!!! PLEASE PROVIDE DETAILED SOLUTIONS AND EXPLANATIONS FOR EACH QUESTION!!!
1. A sewing machine needle moves up and down in simple harmonic motion with an amplitude of 1.27 cm and afrequency of 2.55 Hz.
(a) What is the maximum speed of the needle?
(b) What is the maximum acceleration of the needle?
2. The position of an object that is oscillating on an ideal spring is given by the equation x = (12.3cm) cos[(1.26s-1)t]. At time t = 0.815 s,
(a) how fast is the object moving?
(b) what is the magnitude of the acceleration of the object?
3. A 12.0-N object is oscillating in simple harmonic motion at the end of an ideal vertical spring. Its vertical position y as a function of time t is given by y(t) = 4.50 cm cos[(19.5 s-1)t - /8].
(a) What is the spring constant of the spring?
(b) What is the maximum acceleration of the object?
(c) What is the maximum speed that the object reaches?
(d) How long does it take the object to go from its highest point to its lowest point?
4. An object weighing 44.1 N hangs from a vertical massless ideal spring. When set in vertical motion, the object obeys the equation y(t) = (6.20 cm) cos[(2.74 rad/s)t - 1.40].
(a) Find the time for this object to vibrate one complete cycle.
(b) What are the maximum speed and maximum acceleration of the object.
(c) What is the TOTAL distance the object moves through in one cycle.
(d) Find the maximum kinetic energy of the object.
(e) What is the spring constant of the spring.
5. A lightly damped harmonic oscillator, with a damping force proportional to its speed, is oscillating with an amplitude of 0.500 cm at time t = 0. When t = 8.20 s, the amplitude has died down to 0.400 cm. At what value of t will the oscillations have an amplitude of 0.250 cm?
Explanation / Answer
amplitude, A=1.27 cm
frequency, f=2.55 Hz
a)
V_max=A*W
=A*(2pi*f)
=1.27*10^-2*(2pi*2.55)
=0.203 m/sec
b)
a_max=A*w^2
=A*(2pi*f)^2
=1.27*10^-2*(2pi*2.55)^2
=3.26 m/sec^2
2)
x=(0.123)*cos(1.26*t)
a)
v=dx/dt
v=0.123*1.26*sin(1.26*t)
at t=0.815 sec
v=0.123*1.26*sin(1.26*0.815)
v=2.78*10^-3 m/sec
b)
a=dv/dt
a=-0.123*(1.26)^2*cos(1.26*0.815)
a=-0.195 m/sec^2
magnitude of a=0.195 m/sec^2
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