Review. The top end of a yo-yo string is held stationary. The yo-yo itself is much more massive than the string. It starts from rest and moves down with constant acceleration 0.800 m/s 2 as it unwinds from the string. The rubbing of the string against the edge of the yo-yo excites transverse standing-wave vibrations in the string. Both ends of the string are nodes even as the length of the string increases. Consider the instant 1.20 s after the motion begins from rest. (a) Show that the rate of change with time of the wavelength of the fundamental mode of oscillation is 1.92 m/s. (b) What if? Is the rate of change of the wavelength of the second harmonic also 1.92 m/s at this moment? Explain your answer. (c) What if? The experiment is repeated after more mass has been added to the yo-yo body. The mass distribution is kept the same so that the yo-yo still moves with downward acceleration 0.800 m/s 2 . At the 1.20-s point in this case, is the rate of change of the fundamental wavelength of the string vibration still equal to 1.92 m/s? Explain. (d) Is the rate of change of the second harmonic wavelength the same as in part (b)? Explain.
Review. The top end of a yo-yo string is held stationary. The yo-yo itself is much more massive than the string. It starts from rest and moves down with constant acceleration 0.800 m/s 2 as it unwinds from the string. The rubbing of the string against the edge of the yo-yo excites transverse standing-wave vibrations in the string. Both ends of the string are nodes even as the length of the string increases. Consider the instant 1.20 s after the motion begins from rest. (a) Show that the rate of change with time of the wavelength of the fundamental mode of oscillation is 1.92 m/s. (b) What if? Is the rate of change of the wavelength of the second harmonic also 1.92 m/s at this moment? Explain your answer. (c) What if? The experiment is repeated after more mass has been added to the yo-yo body. The mass distribution is kept the same so that the yo-yo still moves with downward acceleration 0.800 m/s 2 . At the 1.20-s point in this case, is the rate of change of the fundamental wavelength of the string vibration still equal to 1.92 m/s? Explain. (d) Is the rate of change of the second harmonic wavelength the same as in part (b)? Explain.
Review. The top end of a yo-yo string is held stationary. The yo-yo itself is much more massive than the string. It starts from rest and moves down with constant acceleration 0.800 m/s2 as it unwinds from the string. The rubbing of the string against the edge of the yo-yo excites transverse standing-wave vibrations in the string. Both ends of the string are nodes even as the length of the string increases. Consider the instant 1.20 s after the motion begins from rest. (a) Show that the rate of change with time of the wavelength of the fundamental mode of oscillation is 1.92 m/s. (b) What if? Is the rate of change of the wavelength of the second harmonic also 1.92 m/s at this moment? Explain your answer. (c) What if? The experiment is repeated after more mass has been added to the yo-yo body. The mass distribution is kept the same so that the yo-yo still moves with downward acceleration 0.800 m/s2. At the 1.20-s point in this case, is the rate of change of the fundamental wavelength of the string vibration still equal to 1.92 m/s? Explain. (d) Is the rate of change of the second harmonic wavelength the same as in part (b)? Explain.
19:39 ·
C
Chegg
1 69%
✓
The compound beam is fixed at Ę and supported by rollers at A and B. There are pins at C and D. Take
F=1700 lb. (Figure 1)
Figure
800 lb
||-5-
F
600 lb
بتا
D
E
C
BO
10 ft 5 ft 4 ft-—— 6 ft — 5 ft-
Solved Part A The compound
beam is fixed at E and...
Hình ảnh có thể có bản quyền. Tìm hiểu thêm
Problem
A-12
% Chia sẻ
kip
800 lb
Truy cập )
D Lưu
of
C
600 lb
|-sa+ 10ft 5ft 4ft6ft
D
E
5 ft-
Trying
Cheaa
Những kết quả này có
hữu ích không?
There are pins at C and D To F-1200 Egue!)
Chegg
Solved The compound b...
Có Không ☑
|||
Chegg
10
וח
No chatgpt pls will upvote
No chatgpt pls will upvote
Chapter 18 Solutions
Physics for Scientists and Engineers, Technology Update (No access codes included)
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