I am currently studying Simple Harmonicmotion I have a take home assigment that
ID: 1738160 • Letter: I
Question
I am currently studying Simple Harmonicmotion
I have a take home assigment that is asking me to write anexpression for a spring in S.H.M.
Q.)Im requried to use the mass of thespring .25kg and a spring constant of 4.0N/m. The spring isstretched and released and it is found that the velocity of themass is -.174m/s and the acceleration of the mass is .877m/s^2 attime t=.15sec.
Can you help me write an expression for the position of he mass asa function of time. using these given equations.
A-amplitude of an ocillation is the magnitude of the particlesmaximum displacement form he equilibrium position
w-angular frequency
-phase angle
t-time
k-spring constant
f-fequency of an oscillation number of cycles per unit time
f=1/T
w=(k/m)
Force=Mass*acceleration
F=ma
Spring constant * distance from orgin
F=-k(x)
postion
x(t)=Acos(wt-)
veloctiy
v(t)=-wAsin(wt-)
acceleration
a(t)=w^2Acos(wt-)
So far i have a couple of approaches butwhen i back substitute i dont get the given numbers in stated inthe question. For example, i divided angular aceleration byvelcoity and solved for . Placed into the velocityequaion sloved for A(amplitude). Then tryed checking the answer bysloving for postion by using x(t)=Acos(wt+) with sloved valuesfor A and and F=-kx with given values and compared the xvalues from both expressions. Values my values are off, can youhelp me? Anything helps and will be greatlyappreciated!
Explanation / Answer
Let me start with what you have got: Position, x(t)=Acos(wt-) ----------(1)veloctiy
v(t)=-wAsin(wt-)---------(2)
acceleration
a(t)=w^2Acos(wt-)----------(3) For simplicity, put (wt-) = --(4) Dividing (2)/(3) we get, v/a =(1/)Tan ---------(5) Given v = -0.17 m/s and a = 0.877 /s2 at t = 0.15 s Also, = (k/m) = (4/0.25) = 16 = 4rad/s t = 4x0.15 = 0.6 Eqn (5) becomes, -0.17/0.877 = (1/4) Tan Tan = -0.77537 And, = -37.789 degrees = -0.65954rad. Sin = -0.61275 and Cos = 0.79027 From eq(4), = t - = 0.6 - - 0.65954 = 1.25954 = 1.259 rad To deterine amplitude A use eq(2) and substitute the values ofv, and sin A = v/( sin ) = -0.17/(4*-0.61275) = 0.06936 m = 0.0694m We can alsoget the sae using eq(3). Hence, the required expression for the peosion of the particlein SH, x = A cos(t-) x= 0.0694 Cos(4t-1.259) in meters. Note: 1. PL pl use a 0 before decimal fractions; m = 0.25 kg,not .25 kg. This willhelp make the decimal point clear & visible. 2. Study thissolution carefully. ake sure you understand every step. To deterine amplitude A use eq(2) and substitute the values ofv, and sin A = v/( sin ) = -0.17/(4*-0.61275) = 0.06936 m = 0.0694m We can alsoget the sae using eq(3). Hence, the required expression for the peosion of the particlein SH, x = A cos(t-) x= 0.0694 Cos(4t-1.259) in meters. Note: 1. PL pl use a 0 before decimal fractions; m = 0.25 kg,not .25 kg. This willhelp make the decimal point clear & visible. 2. Study thissolution carefully. ake sure you understand every step.
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