- #1
Tangeton
- 62
- 0
For the first part of the equation, I worked out k constant of a string which extended 12.6cm when a mass of 300g was suspended from it using the equation F = kx where F is the force applied and x its extension.
So mg = xk, 300 x 9,81 = 0.126k, the answer was 23357N/m (23kN/m).
The question then says that the string is pulled down by 8.0cm and released, and so it starts oscillating. It then asks me about the frequency that the mass oscillates and the maximum acceleration of the mass.
For the frequency, T = 1/f = 2(pi) * sqrt of m/k , and so f = 1/(2(pi) * sqrt of m/k). My problem is that I been given an extension (8cm) but I already got one k constant, so it doesn't seem to me that I need to work out another k constant because isn't k constant only dependent on the real length of the string, not if the length os extended for oscillations? I guess what I am asking is what would be the k constant in this case: the previous k constant (23kN/m) or the new k constant worked out using 8cm instead of 12.6cm?
So mg = xk, 300 x 9,81 = 0.126k, the answer was 23357N/m (23kN/m).
The question then says that the string is pulled down by 8.0cm and released, and so it starts oscillating. It then asks me about the frequency that the mass oscillates and the maximum acceleration of the mass.
For the frequency, T = 1/f = 2(pi) * sqrt of m/k , and so f = 1/(2(pi) * sqrt of m/k). My problem is that I been given an extension (8cm) but I already got one k constant, so it doesn't seem to me that I need to work out another k constant because isn't k constant only dependent on the real length of the string, not if the length os extended for oscillations? I guess what I am asking is what would be the k constant in this case: the previous k constant (23kN/m) or the new k constant worked out using 8cm instead of 12.6cm?
