Answer:
The positive value of k is 3.
Step-by-step explanation:
The given function is
[tex]y=\cos (kt)[/tex]
Differentiate with respect to t.
[tex]\dfrac{dy}{dt}=\dfrac{d}{dt}\cos (kt)[/tex]
[tex]\dfrac{dy}{dt}=-k\sin (kt)[/tex]
Differentiate with respect to t.
[tex]\dfrac{d^2y}{dt^2}=\dfrac{d}{dt}(-k\sin (kt))[/tex]
[tex]\dfrac{d^2y}{dt^2}=-k\dfrac{d}{dt}(\sin (kt))[/tex]
[tex]\dfrac{d^2y}{dt^2}=-k(k\cos (kt))[/tex]
[tex]\dfrac{d^2y}{dt^2}=-k^2y[/tex] [tex][\because y=\cos (kt)][/tex]
[tex]\dfrac{d^2y}{dt^2}+k^2y=0[/tex] .... (1)
It is given that
[tex]\dfrac{d^2y}{dt^2}+9y=0[/tex] .... (2)
On comparing (1) and (2) we get
[tex]k^2=9[/tex]
[tex]k=\pm\sqrt{9}[/tex]
[tex]k=\pm 3[/tex]
Therefore, the positive value of k is 3.
