Masses m_(A) and m_(B) hanging from the ends of strings of lengths l_(A) and l_(B) are executing. Simple harmonic motions. If their frequencies are related as f_(A)=2f_(B), then……………

Masses m_(A) and m_(B) hanging from the ends of strings of lengths l_(

1.	Masses m_(A) and m_(B) hanging from the ends of strings of lengths l_(A) and l_(B) are executing. Simple harmonic motions. If their frequencies are related as f_(A)=2f_(B), then……………
Answer» `l_(A)=2l_(B) and m_(A)=m_(B)?` `l_(A)=4l_(B)` REGARDLESS of MASSES. `l_(A)=l_(B)//4` regardless of masses `l_(A)=2l_(B)andm_(A)=2m_(B)` Solution :`f=(1)/(2pi)SQRT((G)/(l))`