Answer:
[tex]x(t)=1.5cos(44.27t)cm\\[/tex]
Explanation:
#Using Hooke's Law show that,
[tex]F=kx \ \ \ \ \ \ \ \...eqtn1[/tex]
#And, Newton's Law,
[tex]F=mg \ \ \ \ \ \ \ \ ...eqtn2[/tex]
[tex]F[/tex] is the the force, [tex]k[/tex] is the spring constant and [tex]x[/tex] the compression length.
#Equating [tex]eqtn1 \ and \ eqtn2[/tex] we find:
[tex]kx=mg\\\frac{k}{m}=\frac{g}{x}\\\omega= \sqrt{\frac{k}{m}}=\sqrt{\frac{g}{x}}\\\\\omega ^2=\frac{980cm/s^2}{(50.5cm-50cm}\\\\\omega ^2=1960cm[/tex]
#Using differential equation:
[tex]x\prime \prime +\omega ^2x=0\\x\prime \prime+1960cm(x)=0[/tex] #Solve this differential equation
[tex]x=e^m^y\\m^2e^m^y+(1960cm)e^m^y=0\\m^2+1960cm=0\\m^2=\pm 44.27i[/tex]
#The general solution for this equation is then given as:
[tex]x(t)=c_1sin(44.27t)+c_2cos(44.27t)\\[/tex]
#Using the boundary conditions, at [tex]t=0,x(0)=1.5\\[/tex]
[tex]1.5=c_1sin(0)+c_2cos(0)\\c_2=1.5[/tex]
[tex]=>x(t)=c_1sin(44.27t)+1.5cos(44.27t)[/tex]
Hence displacement at t>0 is given by [tex]x(t)=1.5cos(44.27t)cm\\[/tex]
