SAI'S SCIENCE AND MATHS CLASSES

A mass m is attached to a spring of spring constant K1 If the spring is cut in to two halves and one of the pieces is attached to the same mass m compare the frequencies of oscillation ( Kendriya vidyalaya 2013 ) υ1=1/2π × SQRT(K1/m) Note that the spring constant ∞ (1/length of spring) or K × l = a constant υ2=1/2π × SQRT(2K1/m) ( as length is halved spring constant will be   doubled ) • • • υ1/ υ2=1: √2 Prepared by Sai’s Classes Poonkunnam Thrissur +919846088761

Find the accelaration of block and force of friction   (figure) if coeficent of dynamic friction=0.1 The forces acting on the blocks are obtained by resolving the 20√2 N in to horizontal and vertical components F H =20√2 cos45 =20 (in the direction of motion) F V =20√2 sin45 =20( in the direction of reaction) Weight of block=6kg=60N Frictional force f k =μR=0.1×(60- F V )=0.1×(60-20)= 4N so the net force which causes the accelaration= F H - 4= 20 - 4 =16N accelaration =Net force / mass= 16/6= 8/3 m/sec 2 Prepared by Sai’s Classes Poonkunnam Thrissur +919846088761