The string is now replaced by a spring of spring constant k and natural length l. Mass 2m is fixed at the bottom of the frame . The mass m which has the other endof the spring attached to it is brought near the mass 2m and released as shown in figure. The maximum angle theta that the spring will substend at the centre will be : ( Take k=10N//m,l=1m,m=1kg and l=r)

The string is now replaced by a spring of spring constant k and natura

1.	The string is now replaced by a spring of spring constant k and natural length l. Mass 2m is fixed at the bottom of the frame . The mass m which has the other endof the spring attached to it is brought near the mass 2m and released as shown in figure. The maximum angle theta that the spring will substend at the centre will be : ( Take k=10N//m,l=1m,m=1kg and l=r)
Answer» `60^(@)` `30^(@)` `90^(@)` None of these Solution :Length of spring at maximum `=2l cos phi` `:.` Extension is `x=(2l cos phi -l)` Now initial potential energy of the spring is CONVERTED into FINAL `PE` of spring and gravitational `PE`. `:. (1)/(2)kl^(2)=(1)/(2)k(2lcos theta -l)^(2)+MG(l-lcos theta )` Putting values `(1)/(2)xx10xx1^(2)=(1)/(2)xx10(2 cos phi-1)^(2)+10(1-cos theta)` `:' theta =pi-2phi` `5=5(2 cos phi -1)^(2)+10(1+cos 2phi)` `1=(4 cos ^(2) phi +1)1-4 cos phi)+2 (1+2 cos ^(2)phi-1)` `8 cos ^(2) phi =4 cos phi` `:. cos phi =(1)/(2)` `:. phi =60^(@)` `theta=60^(@)`.