A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes 5T/3. What is the ratio m/M?

A mass M is suspended from a spring of negligible mass. The spring is

1.	A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes 5T/3. What is the ratio m/M?
Answer» Solution :With mass M, the TIME period of the SPRING is `T=2pisqrt((M)/(k))` With mass `M+m`, the time period becomes `(5T)/(3)=2pisqrt((M+m)/(k))or(5)/(3)xx2pisqrt((M)/(k))=2pisqrt((M+m)/(k))` `(25)/(9)M=M+m,(16)/(9)M=mor(m)/(M)=(16)/(9)`