A 0.240 kg block is attached to a spring with a spring constant k = 85.0 N/m. It is set in motion by being displaced 4.00 cm downward. What is the period of its motion? What is its frequency?

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Chapter1: Units, Trigonometry. And Vectors
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A 0.240 kg block is attached to a spring with a spring constant k = 85.0 N/m. It is set in motion by being displaced 4.00 cm downward. What is the period of its motion? What is its frequency? (In the two image are two possibles solution)
EXAMPLE 12-5
Analyzing Waves
Figure 12-27a shows a wave graph, and Figure 12-27b shows a vibration
graph for the same wave. Find the wave's (a) amplitude (A), (b) wavelength
(A), (c) frequency (f). (d) period (T), and (e) speed (v).
Solution:
a. Amplitude:
A = 2(dpeak - duough) = [(+10 cm)-(-10 cm)]
A = 10 cm
%3D
b. Wavelength: A = 20 cm (from Figure 12-27a)
c. Frequency:
f%3D
n cycles 1 cycle
At
2s
(from Figure 12-27b)
%3D
f = 0.5 s
d. Period:
T= 1/f=
0.5 s
%3D
T 2 s
e. Speed:
v = Af (0.2 m)(0.5 s)
v 0.1 m/s
Transcribed Image Text:EXAMPLE 12-5 Analyzing Waves Figure 12-27a shows a wave graph, and Figure 12-27b shows a vibration graph for the same wave. Find the wave's (a) amplitude (A), (b) wavelength (A), (c) frequency (f). (d) period (T), and (e) speed (v). Solution: a. Amplitude: A = 2(dpeak - duough) = [(+10 cm)-(-10 cm)] A = 10 cm %3D b. Wavelength: A = 20 cm (from Figure 12-27a) c. Frequency: f%3D n cycles 1 cycle At 2s (from Figure 12-27b) %3D f = 0.5 s d. Period: T= 1/f= 0.5 s %3D T 2 s e. Speed: v = Af (0.2 m)(0.5 s) v 0.1 m/s
EXAMPLE 12-1
SHM in a Spring-Mass System
A 0.190 kg mass is hung from an ideal spring with a spring constant
k 76.0 N/m. It is set in motion by displacing it 3.00 cm downward, then
releasing it. (a) What is the period of its motion? (b) What is the
frequency?
Solution:
a. Use Equation 12.6 to solve for period:
T= 2TTK
0.190 kg
76.0 N/m
=2T
kg
(kg-m.)/(m-s)
(unit cancellation)
T 0.3141.
A,
T 0.314 s
b. Take the reciprocal of the period to find the frequency:
f==0.3141 s
f= 3.183 s = 3.18 Hz
Notice that the initial displacement, or amplitude, of the mass has no
effect on period (T) or frequency (f).
Transcribed Image Text:EXAMPLE 12-1 SHM in a Spring-Mass System A 0.190 kg mass is hung from an ideal spring with a spring constant k 76.0 N/m. It is set in motion by displacing it 3.00 cm downward, then releasing it. (a) What is the period of its motion? (b) What is the frequency? Solution: a. Use Equation 12.6 to solve for period: T= 2TTK 0.190 kg 76.0 N/m =2T kg (kg-m.)/(m-s) (unit cancellation) T 0.3141. A, T 0.314 s b. Take the reciprocal of the period to find the frequency: f==0.3141 s f= 3.183 s = 3.18 Hz Notice that the initial displacement, or amplitude, of the mass has no effect on period (T) or frequency (f).
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