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 end of 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)`

Options

A. `60^(@)`
B. `30^(@)`
C. `90^(@)`
D. None of these

Answer

A

Explanation

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^(@)`.