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 text5T/3, then the ratio of (m/M) is

Question

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 text5T/3, then the ratio of (m/M) is (A)  (3/5)  (B)  (25/9)  (C)  (16/9)  (D)  (5/3)

Tardigrade Admin · Accepted Answer

T=2π √(M/k) T=2π √(M+m/k) ⇒ (5T/3)=2π √(M+m/k) Dividing Eq. (i) by Eq (ii), we have (3/5)=√(M/M+m) (9/25)=(M/M+m) ⇒ 9M+9m=25M ⇒ 16M=9m (m/M)=(16/9)

Q. 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 $5T/3,$ then the ratio of $\frac{m}{M}$ is

$\frac{3}{5}$

$\frac{25}{9}$

$\frac{16}{9}$

$\frac{5}{3}$

$T = 2 π \frac{M}{k}$ $T = 2 π \frac{M + m}{k}$ $\Rightarrow$ $\frac{5 T}{3} = 2 π \frac{M + m}{k}$ Dividing Eq. (i) by Eq (ii), we have $\frac{3}{5} = \frac{M}{M + m}$ $\frac{9}{25} = \frac{M}{M + m}$ $\Rightarrow$ $9 M + 9 m = 25 M$ $\Rightarrow$ $16 M = 9 m$ $\frac{m}{M} = \frac{16}{9}$

Q. 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, then the ratio of Mm​ is

53​

925​

916​

35​