(a) A device consists of an object with a weight of 25.0 N hanging vertically from a spring with a spring constant of 250 N/m. There is negligible damping of the oscillating system. Applied to the system is a harmonic driving force of 13.0 Hz, which causes the object to oscillate with an amplitude of 4.00 cm. What is the maximum value of the driving force (in N)? (Enter the magnitude.) 670.8 N (b) What If? The device is altered so that there is a damping coefficient of b = 5.00 N s/m. The hanging weight and spring constant remain the same. The same driving force as found in part (a) is applied with the same frequency. What is the new amplitude (in cm) of oscillation? cm (c) What If? Repeat the same calculation as part (b), only now with a damping coefficient of b = 100 N s/m. (Enter the answer in cm.) cm

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(a) A device consists of an object with a weight of 40.0 N hanging vertically from a spring with a spring constant of 250 N/m. There
is negligible damping of the oscillating system. Applied to the system is a harmonic driving force of 11.0 Hz, which causes the
object to oscillate with an amplitude of 4.00 cm. What is the maximum value of the driving force (in N)? (Enter the magnitude.)
769.11 N ✓
(b) What If? The device is altered so that there is a damping coefficient of b = 5.00 N s/m. The hanging weight and spring
constant remain the same. The same driving force as found in part (a) is applied with the same frequency. What is the new
amplitude (in cm) of oscillation?
6.523 cm. x
Enter a number.
(c) What If? Repeat the same calculation as part (b), only now with damping coefficient of b = 100 N s/m. (Enter the answer in
cm.)
Show transcribed image text
Answers
narendra answered this
1,701 answers
Solution:
Given That
m=
40
g
201 - من
2
= 2πT XII = 69.11 rad|s
x= no Sin wt
No = 4cm
amplitude Xo
Here
501
что =4.08 ку
9.8
= 0.04 m
folk
√(1-9) 7(269) ²
7
Fo=
q
f
6= 69-11
حبا
соп
O
con =
Was this answer helpful?
250×9.8
m
40
wn = 7.8262 radls
q=
69.11 = 8.83
7.8261
Xo (9²-1) k
Fo=769.689 N
k=250N/m
m
TS
= 0.04 (8.83²-1) (250)
5
F = Fosin t
0
View
Ⓒ
C= 5N-1/m
Xo =
Xo
f =
f
Xo =
f=
=
f:
Xo = 0.0399
Хо
Xo =
√(1-9²1²+(279)²
5
Xo = 4cm
Xo =
Zoom
2X408x7.8262
0.07829
769.689/250
√(1-(8.833²)² + (2x0.07829x8.83) ²
769.689/250
76.9813
3.99 = 4 Cm
Share Highlight
100
2x4.08 x 7.8262
1.5658
(cc = 2mwn)
769,689/250
(1-8.83²)² + (2x1.5658X8,833²
769689/250
81.7853
Xo = 3.764 cm
= 0.03764
Transcribed Image Text:(a) A device consists of an object with a weight of 40.0 N hanging vertically from a spring with a spring constant of 250 N/m. There is negligible damping of the oscillating system. Applied to the system is a harmonic driving force of 11.0 Hz, which causes the object to oscillate with an amplitude of 4.00 cm. What is the maximum value of the driving force (in N)? (Enter the magnitude.) 769.11 N ✓ (b) What If? The device is altered so that there is a damping coefficient of b = 5.00 N s/m. The hanging weight and spring constant remain the same. The same driving force as found in part (a) is applied with the same frequency. What is the new amplitude (in cm) of oscillation? 6.523 cm. x Enter a number. (c) What If? Repeat the same calculation as part (b), only now with damping coefficient of b = 100 N s/m. (Enter the answer in cm.) Show transcribed image text Answers narendra answered this 1,701 answers Solution: Given That m= 40 g 201 - من 2 = 2πT XII = 69.11 rad|s x= no Sin wt No = 4cm amplitude Xo Here 501 что =4.08 ку 9.8 = 0.04 m folk √(1-9) 7(269) ² 7 Fo= q f 6= 69-11 حبا соп O con = Was this answer helpful? 250×9.8 m 40 wn = 7.8262 radls q= 69.11 = 8.83 7.8261 Xo (9²-1) k Fo=769.689 N k=250N/m m TS = 0.04 (8.83²-1) (250) 5 F = Fosin t 0 View Ⓒ C= 5N-1/m Xo = Xo f = f Xo = f= = f: Xo = 0.0399 Хо Xo = √(1-9²1²+(279)² 5 Xo = 4cm Xo = Zoom 2X408x7.8262 0.07829 769.689/250 √(1-(8.833²)² + (2x0.07829x8.83) ² 769.689/250 76.9813 3.99 = 4 Cm Share Highlight 100 2x4.08 x 7.8262 1.5658 (cc = 2mwn) 769,689/250 (1-8.83²)² + (2x1.5658X8,833² 769689/250 81.7853 Xo = 3.764 cm = 0.03764
(a) A device consists of an object with a weight of 25.0 N hanging vertically from a spring with a spring constant of 250 N/m. There is negligible damping of the oscillating system.
Applied to the system is a harmonic driving force of 13.0 Hz, which causes the object to oscillate with an amplitude of 4.00 cm. What is the maximum value of the driving force (in
N)? (Enter the magnitude.)
670.8
N
(b) What If? The device is altered so that there is a damping coefficient of b = 5.00 N. s/m. The hanging weight and spring constant remain the same. The same driving force as found
in part (a) is applied with the same frequency. What is the new amplitude (in cm) of oscillation?
cm
(c) What If? Repeat the same calculation as part (b), only now with a damping coefficient of b = 100 N s/m. (Enter the answer in cm.)
cm
Transcribed Image Text:(a) A device consists of an object with a weight of 25.0 N hanging vertically from a spring with a spring constant of 250 N/m. There is negligible damping of the oscillating system. Applied to the system is a harmonic driving force of 13.0 Hz, which causes the object to oscillate with an amplitude of 4.00 cm. What is the maximum value of the driving force (in N)? (Enter the magnitude.) 670.8 N (b) What If? The device is altered so that there is a damping coefficient of b = 5.00 N. s/m. The hanging weight and spring constant remain the same. The same driving force as found in part (a) is applied with the same frequency. What is the new amplitude (in cm) of oscillation? cm (c) What If? Repeat the same calculation as part (b), only now with a damping coefficient of b = 100 N s/m. (Enter the answer in cm.) cm
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