Orders of Magnitude and Correct Answer

emmanuel (ae8272) – hw07-Waves and vibrations – dickerson – (28763)
This print-out should have 47 questions.
Multiple-choice questions may continue on the next column or page – find all choices before answering.

1

004 (part 4 of 6) 10.0 points
What is its frequency?
Correct answer: 6.1216 Hz.

001 (part 1 of 6) 10.0 points
A spring with a spring constant of 50.3 N/m is attached to different masses, and the system is set in motion.
What is its period for a mass of 2.8 kg?

1
= 6.1216 Hz .
0.163356 s

f2 =

005 (part 5 of 6) 10.0 points
What is the period for a mass of 0.65 kg?

Correct answer: 1.48243 s.
Explanation:
Let :

Explanation:

Correct answer: 0.714254 s.

k = 50.3 N/m m1 = 2.8 kg .

and

T = 2π

m k T1 = 2 π

2.8 kg
= 1.48243 s .
50.3 N/m

002 (part 2 of 6) 10.0 points
What is its frequency?

Explanation:
Let :

T3 = 2 π

m3 = 0.65 kg .

0.65 kg
= 0.714254 s .
50.3 N/m

006 (part 6 of 6) 10.0 points
What is its frequency?
Correct answer: 1.40006 Hz.
Explanation:

Correct answer: 0.674567 Hz.
Explanation:
f= f1 =

f3 =

1
T

1
= 0.674567 Hz .
1.48243 s

003 (part 3 of 6) 10.0 points
What is the period for a mass of 34 g?
Correct answer: 0.163356 s.
Explanation:

007 10.0 points
A 51 g object is attached to a horizontal spring with a spring constant of 14.4 N/m and released from rest with an amplitude of
27.2 cm.
What is the velocity of the object when it is halfway to the equilibrium position if the surface is frictionless?
Correct answer: 3.95818 m/s.

Let : m2 = 34 g = 0.034 kg .

T2 = 2 π

1
= 1.40006 Hz .
0.714254 s

0.034 kg
= 0.163356 s .
50.3 N/m

Explanation:
Let : m = 51 g = 0.051 kg , k = 14.4 N/m , and
A = 27.2 cm = 0.272 m .

emmanuel (ae8272) – hw07-Waves and vibrations – dickerson – (28763) directed toward the equilibrium position.

The speed is v= =

2

k 2
(A − x2 ) m 14.4 N/m
[(0.272 m)2 − (0.136 m)2 ]
0.051