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Old Jun 25th 2014, 02:26 AM   #1
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Amplitude of string when the string breaks

the period will be shorter due to smaller mass of oscillating system... but the amplitude is greater( ans from my book , in my opinion, the amplitude should be smaller because the string is (dispalcement) is displaced lesser if the mass is removed
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Old Jun 25th 2014, 08:07 AM   #2
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You are correct that the period decreases, due to loss of mass. As for the amplitude: I think this is a bit subtle. When the mass is first suspended from the end of the metal strip, the strip deflects a certain amount under its weight. Let's say that initial deflection is y= -D. Then when you pull down further on the mass and set it vibrating the amplitude of vibration is A, as measured from point D. In other words the mass vibrates about point D with amplitude A: y = -D + Asin(omega t), reaching max negative displacement of y=-(A+D) and max positive displacement of y=A-D. When the string is cut the metal strip is at y=-(A+D) and its center point of vibration is now at y=0. In other words there is no more offset of D, and the metal strip vibrates about a point that is in line with the table - hence its amplitude is of vibration is now A+D. Hence the amplitude of vibration has increased.

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Old Jun 25th 2014, 10:36 PM   #3
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Originally Posted by ChipB View Post
You are correct that the period decreases, due to loss of mass. As for the amplitude: I think this is a bit subtle. When the mass is first suspended from the end of the metal strip, the strip deflects a certain amount under its weight. Let's say that initial deflection is y= -D. Then when you pull down further on the mass and set it vibrating the amplitude of vibration is A, as measured from point D. In other words the mass vibrates about point D with amplitude A: y = -D + Asin(omega t), reaching max negative displacement of y=-(A+D) and max positive displacement of y=A-D. When the string is cut the metal strip is at y=-(A+D) and its center point of vibration is now at y=0. In other words there is no more offset of D, and the metal strip vibrates about a point that is in line with the table - hence its amplitude is of vibration is now A+D. Hence the amplitude of vibration has increased.

since you said that there's no offset of D , then why the ampltiude is now y=A+D ? in my opinion it it's now the string oscillates at y=o, which positive displacement is y=A , and negative displacement is y= -A ..
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Old Jun 26th 2014, 05:49 AM   #4
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After the string is cut the metal strip oscillates about the y=0 line. It's initial displacement from y=0 (at the instant the string breaks) is -(A+D). Hence its amplitude of oscillation will be A+D. See the attached figure.
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Old Jun 26th 2014, 10:26 AM   #5
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Originally Posted by ChipB View Post
After the string is cut the metal strip oscillates about the y=0 line. It's initial displacement from y=0 (at the instant the string breaks) is -(A+D). Hence its amplitude of oscillation will be A+D. See the attached figure.
wow! thank you! i can understand it finally.... it's really hard to understand it.
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Old Jun 26th 2014, 10:05 PM   #6
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Originally Posted by ChipB View Post
After the string is cut the metal strip oscillates about the y=0 line. It's initial displacement from y=0 (at the instant the string breaks) is -(A+D). Hence its amplitude of oscillation will be A+D. See the attached figure.

sorry, based on your explaination, the amplitude is measured from the point y=0. during the string is attached with the mass( string is displaced) , so you have positive displacement y = -D + Asin(omega t), ..

my question is since the string now is at y=-D, why not the amplitude measured from the y=-D , but at y=0?
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Old Jun 27th 2014, 04:33 AM   #7
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Originally Posted by ling233 View Post
, based on your explaination, the amplitude is measured from the point y=0. during the string is attached with the mass( string is displaced) , so you have positive displacement y = -D + Asin(omega t), ..

my question is since the string now is at y=-D, why not the amplitude measured from the y=-D , but at y=0?
In the begining when the string and mass are still attached to the virbrating metal strip, the amplitude of vibration is indeed measured from the point y=-D, because that's the equilibrium point for the system. See the top half of the diagram I posted earlier - note how the amplitude A is measured.

Once the string breaks and the mass is no longer attached to the vibrating metal strip the equilibrium point is now at y = 0, and the amplitude is A' as shown in the bottom half of the diagram.
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Old Jun 27th 2014, 06:21 AM   #8
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Exclamation

Originally Posted by ChipB View Post
In the begining when the string and mass are still attached to the virbrating metal strip, the amplitude of vibration is indeed measured from the point y=-D, because that's the equilibrium point for the system. See the top half of the diagram I posted earlier - note how the amplitude A is measured.

Once the string breaks and the mass is no longer attached to the vibrating metal strip the equilibrium point is now at y = 0, and the amplitude is A' as shown in the bottom half of the diagram.
for the bottom diagram, amplitde is measured from y=0, then the amplitude should be =A am i right? since amplitude is the max displacement form the equuilibrium point (y=0) ,
for the upper diagram , the string vibrate about y=-D.. so the amplitude A= is the distance from the y=-D to the max displacement from y=-D ...

for both cases ampltide are the same.... please correct my concept ..
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Old Jun 27th 2014, 08:11 AM   #9
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Originally Posted by ling233 View Post
for the bottom diagram, amplitde is measured from y=0, then the amplitude should be =A am i right?
No - in the lower diagram the amplituide is A' ("A prime") - which is of greater magnitude than A in the upper diagram.
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Old Jun 27th 2014, 08:57 AM   #10
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Originally Posted by ChipB View Post
No - in the lower diagram the amplituide is A' ("A prime") - which is of greater magnitude than A in the upper diagram.
why the A' shouldnt be equal to A? why you do it in y=A+D , which is the distance between the equilibrium position when the mass is put on the string, and the amplitude.....

the question only ask for amplitude am i right?
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