Soal dan Pembahasan Gerak Harmonik Sederhana
Simple
Harmonic Motion
1. Two
springs with constants respectively equal to 150 N/m arranged in parallel as
shown in the following figure. What is the period of structure ( the mass of
object m is 3 kg)?
A.
0,2 π s

B. 0,4
π s
C. 2,0
π s
D. 3,0
π s
E. 4,0
π s
2. An
object is vibrating until form a harmonic motion with an equation y = 0,04 sin 20π t. The deviation of y is in meters, t
is time in second. What is the amplitude and frequency?
A. 0,04 m and 5 Hz
B. 0,04 m and 10 Hz
C. 0,4 m and 10 Hz
D. 0,5 m and 5 Hz
E. 4 m and 5 Hz
3. A
mass of block is 0,5 kg, connected to a
light spring with constant 200 N/m. Then the system is oscillating
harmonic. If we know the maximum deviation is 3 cm, the maximum speed is. . .
A. 0,1 m/s
B. 0,6 m/s
C. 1 m/s
D. 1,5 m/s
E. 2 m/s
4.
Two identical spring
with equal constants of 200 N/m are arranged in series as shown in the
following figure. Object m is 2 kg suspended on the lower end of the string.
The period of the string is. . .

A. 0,2
π √2 s
B. 0,2
π √3 s
C. 2,0
π √2 s
D. 2,0
π √3 s
E. 3,0
π √2 s
5. The
mass of an object with moves in a harmonic is 400 kg at a frequency of 25 Hz
and amplitude of 5 cm. What is the maximum acceleration of harmonic motion?
A. 5000π2 cm/s2
B. 500π2 cm/s2
C. 10800π2 cm/s2
D. 18570 π2 cm/s2
E. 18750π2 cm/s2
KEY
ANSWER :
1. A.
0,2 π s
2. B.
0,04 m and 10 Hz
3. B.
0,6 m/s
4. A.
0,2 π √2 s
5. E. 18750π2 cm/s2
SOLUTION
:
1. Given : k = 150 N/m
m =
3 kg
Ask : T. . .
Answer : kt =
k1 + k2
= 150 + 150 = 300 N/m
T = 2π √m/kt
=
2π √3/300
= 2π/10 = 0,2 π s
2. Given : y = 0,04 sin 20π t
Deviation y =
meters
Time =
second
Ask :
amplitude and frequency. . .
Answer :
amplitudo atau A
y = 0,04 sin 20π t
↓
A = 0,04 meter
↓
A = 0,04 meter
frekuensi atau f
y = 0,04 sin 20π t
↓
ω = 20π
2πf = 20π
f = 10 Hz
3. Given : m = 0,5 kg
k
= 200 N/m
ymaks
= A = 3 cm = 0,03 m
Ask : Vmax. . .
Answer : vmaks = ω A
2π
vmaks= ____ x A
T
2π
vmaks = ______ x (0,03) = 0,6 m/s
0,1 π
2π
vmaks= ____ x A
T
2π
vmaks = ______ x (0,03) = 0,6 m/s
0,1 π
4. Given : k = 200 N/m
m = 2 kg
Ask : T. . .
Answer : 1/kt = 1/k1 +1/k2
=
1/200 + 1/200
Kt =
100 N/m
T =
2π √m/kt
=
2π √2/100
=
(2π √2) : 10 = 0,2 π √2 s
5. Given : m = 400
kg
F =
25 Hz
Amplitude = 5 cm
Ask :
maximum acceleration. . .
Answer : a
maks = ώ2 A
=
(2 πf)2 (5)
=
(2π . 25)2.7,5
=
4π2.625.7,5
=
18.750π2 cm/s2
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