Periodic and Circular Motion Periodic and Circular Motion Physics Help Forum Jun 25th 2014, 02:26 AM #1 Senior Member   Join Date: Jun 2014 Posts: 306 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   Jun 25th 2014, 08:07 AM #2 Physics Team   Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,344 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. Last edited by ChipB; Jun 25th 2014 at 08:34 AM.   Jun 25th 2014, 10:36 PM   #3
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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.

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 ..   Jun 26th 2014, 05:49 AM #4 Physics Team   Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,344 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.   Jun 26th 2014, 10:26 AM   #5
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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.
wow! thank you! i can understand it finally.... it's really hard to understand it.   Jun 26th 2014, 10:05 PM   #6
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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.

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?   Jun 27th 2014, 04:33 AM   #7
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 Originally Posted by ling233 , 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.   Jun 27th 2014, 06:21 AM   #8
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Posts: 306 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 ..   Jun 27th 2014, 08:11 AM   #9
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 Originally Posted by ling233 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.   Jun 27th 2014, 08:57 AM   #10
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 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?  Tags amplitude, breaks, string Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post kasia.ogor Waves and Sound 1 Nov 16th 2010 09:23 PM apple123 Waves and Sound 1 Mar 23rd 2010 10:35 PM JoanF Equilibrium and Elasticity 3 Dec 28th 2009 08:54 AM IrrationalPi3 Waves and Sound 1 Oct 8th 2009 05:18 AM cooltowns Periodic and Circular Motion 4 Jun 3rd 2009 11:00 PM