SHM செய்யும் துகளின் முடுக்கம் 3 தூரத்தில் ` 12 cm // sec^(2) ` cm

The acceleration of a particle performing S.H.M. is 12 cm // sec^(2) cm at a distance of 3 cm form the mean position. Its period is

The acceleration of a particle performing SHM is 12 cm //s^(2) at a distance of 3 cm from the mean position . Calculate its time - period . Hint : a = omega^(2) x a = 12 cm//s^(2) , x=3cm Find omega T= (pi)/(omega)

The maximum displacement of a particle executing S.H.M. is 1 cm and the maximum acceleration is (1.57)^(2) cm per sec^(2) . Then the time period is

If the displacement, velocity and acceleration of a particle in SHM are 1 cm, 1cm/sec, 1cm/ sec^2 respectively its time period will be (in seconds) :

A particle is executing S.H.M. from mean position at 5 cm distance, acceleration is 20 cm//sec^(2) then value of angular velocity will be

The area of three consecutive faces of a cuboid are 12 cm^(2) . 20 cm^(2) and 15 cm^(2) , then the volume ( in cm^(3) ) of the cuboid is

The areas of three consecutive faces of a cuboid are 12 cm^(2), 20 cm^(2) and 15 cm^(2) , then the volume (in cm^(3) ) of the cuboid is

Suppose a gas sample in all have 6xx10^(23) molecules Each 1//3rd of the molecules have rms speed 10^(4) cm//sec, 2 xx 10^(4) cm//sec and 3 xx 10^(4) cm//sec Calculate the rms speed of gas molecules in sample .

The maximum velocity and acceleration of a particle in S.H.M. are 100 cms/sec and 157 cm/ sec^2 respectively. The time period in seconds will be :

A particle moves on a line according to the law s=at^(2)+bt+c . If the displacement after 1 sec is 16 cm, the velocity after 2 sec is 24 cm/sec and acceleration is 8cm//sec^(2) , then