- #1
snowcrystal42
Am I thinking about this problem correctly?
"A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of 3 Hz. The amplitude of the motion is 5.08 x 10-2 m. At the point where the block has its maximum speed, it suddenly splits into two parts, only one part remaining attached to the string. (a) What are the amplitude and frequency of the simple harmonic motion that exists after the block splits? (b) What about when the block splits while at one of the extreme ends?"
(a) First of all, the answer key I was given has an equation that looks like it relates to elastic potential energy (it says ½mv2 = ½kA2). Isn't the point where the speed is at its max also where the box is at its equilibrium position (and in this case where the spring is unstretched) so the elastic potential energy is zero? How would elastic potential energy be related to this otherwise? Shouldn't the object/spring only have translational kinetic energy?
Since Vmax = Aω and ω = √(k/m)
A = Vmax√(m/k)
Since the mass is halved, A is divided by √2 so the answer is just the original amplitude divided by √2, right?
Then if Aω = Vmax, A(2πƒ) = Vmax so ƒ = Vmax / (2πA), so if A is divided by √2 then frequency is multiplied by √2, correct?
(b) This is the part that sort of confuses me. At the ends, V = 0 so there is no kinetic energy; all the energy is in the form of potential energy. If PE = ½(kx2), then mass isn't a part of determining the potential energy, so the potential energy stays the same? If the potential energy doesn't change, why does this mean that the amplitude doesn't change?
From a non-mathematical point of view, why does it matter where you break the block in half?
And if the amplitude doesn't change, why does the frequency change? And how do you know by how much? Is it because ω = 2πƒ = √(k/m) so if the mass is halved then the frequency is multiplied by √2 ?
Thanks!
"A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of 3 Hz. The amplitude of the motion is 5.08 x 10-2 m. At the point where the block has its maximum speed, it suddenly splits into two parts, only one part remaining attached to the string. (a) What are the amplitude and frequency of the simple harmonic motion that exists after the block splits? (b) What about when the block splits while at one of the extreme ends?"
(a) First of all, the answer key I was given has an equation that looks like it relates to elastic potential energy (it says ½mv2 = ½kA2). Isn't the point where the speed is at its max also where the box is at its equilibrium position (and in this case where the spring is unstretched) so the elastic potential energy is zero? How would elastic potential energy be related to this otherwise? Shouldn't the object/spring only have translational kinetic energy?
Since Vmax = Aω and ω = √(k/m)
A = Vmax√(m/k)
Since the mass is halved, A is divided by √2 so the answer is just the original amplitude divided by √2, right?
Then if Aω = Vmax, A(2πƒ) = Vmax so ƒ = Vmax / (2πA), so if A is divided by √2 then frequency is multiplied by √2, correct?
(b) This is the part that sort of confuses me. At the ends, V = 0 so there is no kinetic energy; all the energy is in the form of potential energy. If PE = ½(kx2), then mass isn't a part of determining the potential energy, so the potential energy stays the same? If the potential energy doesn't change, why does this mean that the amplitude doesn't change?
From a non-mathematical point of view, why does it matter where you break the block in half?
And if the amplitude doesn't change, why does the frequency change? And how do you know by how much? Is it because ω = 2πƒ = √(k/m) so if the mass is halved then the frequency is multiplied by √2 ?
Thanks!