State of differential equation equation of linear simple harmonic motion. Hence obtain the expression for acceleration, velocity and displacement of a particle performing linear S.H.M.
A body cools from `80^(@)C` to `70^(@)C` in 5 minutes and to `62^(@)C` in the next 5 minutes. Calculate the temperature of the surroundings.

Explanation

The differential equation of linear SHM is `(d^(2) vecx)/(dt^(2))+(k)/(m) vecx=0`
where, m = mass of the particle performing S.H.M.
`(d^(2)vecx)/(dt^(2))=` acceleration of the particle when its displacement from the mean position is `vecx`, and k = froce constant For linear motion, we can write the differential equation in scalar form
`(d^(2)x)/(dt^(2))+omega^(2)x=0`
Let `(k)/(m)=omega^(2), and omega` is angular frequency.
`therefore (d^(2)x)/(dt^(2))+omega^(2)x=0`
`therefore " Acceleration", a=(d^(2)x)/(dt^(2))= -omega^(2)x " " ` ....(i)
The minus sign shows that the acceleration and the displacement have opposite directions.
Aceleration, ` a=(d^(2)x)/(dt^(2))=(dv)/(dt)=(dv)/(dx).(dx)/(dt)`
But, `(dx)/(dt)=v`
Where, v is velocity.
Acceleration, `a=(dv)/(dx)xxv=v(dv)/(dx) " " ` ...(ii)
Substituting equation (ii) in equation (i), we get
`v (dv)/(dx)=-omega^(2)x`
`therefore v dv = -omega^(2) x dx`
Intergrating both sides, we get
`int v dv = int - omega^(2) x dx= - omega^(2) int x dx`
`(v^(2))/(2)=(-omega^(2)x^(2))/(2)+C " " `...(iii)
Where, C is constant of integration.
To find C, consider boundary value condition. At an extreme position ( a turning point of the motion), the velocity of the particle is zero.
Thus v=0 when `x =pmA`, where A is the amplitude.
`therefore 0=(-omega^(2)A^(2))/(2)+C`
`therefore C=(omega^(2)A^(2))/(2)`
From equation (iii),
`(v^(2))/(2)=(-omega^(2)x^(2))/(2)+(omega^(2)A^(2))/(2)`
`therefore v^(2)=omega^(2)(A^(2)-x^(2))`
`therefore v=pm omega sqrt(A^(2)-x^(2)) " " `...(iv)
This equation gives the velocity of the particle in terms of displacement x. The velocity towards right is taken to be positive and towrds left is taken to be negative.
Since, `v=(dx)/(dt)`,
We can write equation (iv) as follows
`(dx)/(dt)=omega sqrt(A^(2)-x^(2))`
(considering only the plus sign)
`therefore (dx)/( sqrt(A^(2)-x^(2)))=omega dt`
On intergrating both sides, we get
`sin^(-1)((x)/(A))=omega t+alpha " " ` ....(v)
Where, the constant of integration `alpha` is found from the initial condition (phase angle).
It is also called epoch.
From equation (v), we have
`((x)/(A))=sin(omega t+alpha)`
`therefore` Displacement as a function of time is `x=A sin(omega +alpha)`
This is an expression for displacement of particle performing S.H.M.
Numerical :
`theta_(1)=(80+70)/(2)=75^(@)C`
`theta_(2)=(70+62)/(2)=66^(@)C`
`(d theta_(1))/(dt)=(80-70)/(5)=2^(@)C//min`
`(d theta_(2))/(dt)=(70-62)/(5)=1.6^(@)C//min`
According to Newton's law of cooling,
`(d theta_(1))/(dt)=K(theta_(1)+theta_(0))`
`and (d theta_(2))/(dt)=K(theta_(2)-theta_(0))`
where, `theta_(0)` is the temperature of surrounding.
`((d theta_(1))/(dt))/((dtheta_(2))/(dt))=(K(theta_(1)-theta_(0)))/(K(theta_(2)-theta_(0)))=((theta_(1)-theta_(0)))/((theta_(2)-theta_(0)))`
`therefore (2)/(1.6)=(75-theta_(0))/(66-theta_(0))`
`2(66-theta_(0))=1.6(75-theta_(0))`
`2(66-theta_(0))=1.6(75-theta_(0))`
`132 - 2 theta_(0)=120-1.6 theta_(0)`
`132 -120= 2theta_(0)-1.6 theta_(0)`
`12=0.4 theta_(0)`
`theta_(0)=(12)/(0.4)=30^(@)C`
`therefore` Temperature of surrounding, `theta_(0)=30^(@)C.`