a) A weight is connected to a spring. When the system is in balance it is still
ID: 1951487 • Letter: A
Question
a)
A weight is connected to a spring. When the system is in balance it is still (not moving).
When the weight is moved from it's balance point and then released (at the time t=0 s) it swings. The location of the weight is descried with the formula:
x(t) = A cos(t).
Where:
A = Amplitude
= angular velocity
T = is the time it takes the weight whole swing from beginning to the end. x(t) = x(t + T)
Use the properties of trigonometry (cos, sin tan). The rules of trigonometry to find the connection between T and . [Hint: Use radians and sequence of trigonometry to find the answer]
b)
Show that the acceleration a(t), is in proportion with the location x(t) a(t) = K x(t) where K is a constant.
Explanation / Answer
Given: x(t) = Acos(wt) Say you have cos(t): its period T is 2pi Say you have cos(2t): its period is pi. In general, the period of: cos(at) is found by doing the following. the period T is the value such that: aT = 2pi, solve for T T = 2pi/a In : x(t) = Acos(wt) wT = 2pi T = 2pi/W that's the relation between the period T and the angular velocity w. For the second part, there are two ways to do this. First way is from principles using Newton's second law and Hooke's law. Newton's 2nd law is: F = ma Hooke's law: F = -kx k is the spring constant, x is the displacement of the mass as it hangs on the spring. Using Newton's law of motion: the mass pulls on the spring due to gravity, and the spring in turns exerts an upward force on the mass to counter gravity. In mathematical terms: F = ma = -kx ma = -kx a = (-k/m)x From your problem description: x(t) = Acos(wt) a(t) = (-k/m)Acos(wt) a(t) = -(k/m)Acos(wt) Letting : K = -k/m a(t) = Kx(t) The second derivation uses calculus, so ignore it if you have not yet encountered derivatives: x(t) = Acos(wt) The velocity function is given by: v(t) = x'(t) = -Awsin(wt) The acceleration function is given by: a(t) =-(w^2)Acos(wt) Letting : -w^2 = K a(t) = Kx(t) Quick note: w^2 = k/m because : w = sqrt(k/m) w^2 = k/m Mostly likely you might not have seen derivatives yet, so only focus on the first derivation which uses first principles.
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