Solution please of 6 and 7 answer of 6 is root2 g/10 m/s^2 and of 7 it is pi(root1 - Physics - Ray Optics And Optical Instruments

Question

Solution please of 6 and 7 answer of 6 is root2 g/10 m/s^2
and of 7 it is pi(root1 2 +1)/rootg = 2 1 sec

Q6 A simple pendulum suspended through the celling of a car rest
oscillates with time period 2 seconds car suddenly starts moving
with acceleration a0 then time period changes to 1 99
seconds Find a0

Q7 A 0 1kg ball is attached to string 1 2 m long and suspended as a
simple pendulum At a point 0 2 m below the point of suspension a
peg is placed which the string hits when the pendulum comes down If
the mass is pulled a small distance to one side and released what
will be the time period of the motion (g = 10 m/sec2)

Vipra Mishra · Accepted Answer

6
Dear Student
When car is rest then time period will be
 Time period T =2πLg2=2πLgL=gπ2
The
car is accelerating in horizontal direction with acceleration a The
acceleration due to gravity is g in downward direction

Resultant acceleration
a' =ao2+g2New Time period T '=2πLa'1
99=2πLa'3
9601=4π2La'Substituting value of L and a'3
9601=4π2gπ2ao2+g20 99=gao2+g20 9801=g2ao2+g20 9801ao2+0
9801g2=g20 9801ao2=0 019g2ao2=0 020g2ao=0 14g

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