Find the period of small oscillations of a mathematical pendulum of length `l` if its point of suspension `O` moves relative to the Earth's surface in an arbitrary directio with a constant acceleration `w` (figure). Calculate that period if `l=21 cm, w=g//2, ` and the angle between the vectors `w` and `g` equals `beat =120^(@)`.

Explanation

In the frame of point of suspension the mathematical pendulum of mass `m` (say) will oscillate. In this fracem, the body `m` will experience the inertial force `m(-vec(w))` in addition to the real forces during its oscillations. Therefore in equilibrium position `m` is deviated by some angle say `alpha`. In equilibrium position.
`T_(0)cos alpha=mg+ m w cos ( pi-beta)` and `T_(0)sin alpha=m w sin (pi - beta)`
So, from these two two Eqns
`tan alpha =(g-w cos beta)/(w sin beta)`
and `cos alpha=sqrt((m^(2)w^(2)sin^(2)beta+(m g - m w cos beta)^(2))/(mg - m w cos beta)) ....(1)`
Let us displace the bob `m` from its equilibrium position by some small angle and then release it. Now locate the ball at an angular position `(alpha + theta)` from vertical as shown in the figure.
From the Eqn. : `N_(0z)=Ibeta_(z)`
`=-mglsin (alpha+theta)-m w cos (pi - beta)l sin (alpha= theta )+ m w sin (pi - beta) l cos ( alpha + theta)=m l^(2) ddot(theta)`
or , `- g(sin alpha cos theta+cos alpha sin theta)- w cos (pi - beta)( sin alpha cos theta+cos alpha sin theta_+ w sin beta ( cos alpha cos theta - sin alpha sin theta)`
`=l ddot(theta)`
But for small
`theta, sin theta ~=theta cos theta ~=1`
So, `-g( sin alpha + cos alpha theta)-w cos (pi - beta)( sin alpha + cos alpha theta)=w sin beta ( cos alpha - sin alpha theta)`
`=lddottheta`
or `(tan alpha+theta)(w cos beta-g)+w sin beta( 1- tan alpha theta)=(l)/(cos alpha)ddot(theta) ...(2)`
Solving Eqns `(1)` and `(2)` simultaneously we get
`-(g^(2)-2 w g cos beta + w ^(2)) theta=l sqrt(g^(2)+w^(2)-2 w g cos beta) ddot ( theta)`
Thus `ddot(theta)=-(|vec(g)-vec(w)|)/(l) theta`
Hence the sought time period `T=(2pi)/(omega_(0))=2pi sqrt((l)/(|vec(g)-vec(w)|))`