A body of mass m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass when the mass m is slightly pulled down and released, it oscillates with a time period of 3 s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5 s. The value of m in kg is

Question

A body of mass m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass when the mass m is slightly pulled down and released, it oscillates with a time period of 3 s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5 s. The value of m in kg is (A) (3/4) (B) (4/3) (C) (16/9) (D) (9/16)

Tardigrade Admin · Accepted Answer

3 = 2π √(m/k) 5=2π√(m+1/k) (3/5) = √(m/m+1) (9/25) = (m/m+1) 9m+9=25m 16m = 9 m = (9/16)

$\frac{3}{4}$

$\frac{4}{3}$

$\frac{16}{9}$

$\frac{9}{16}$

$3 = 2 π \frac{m}{k}$
$5 = 2 π \frac{m + 1}{k}$
$\frac{3}{5} = \frac{m}{m + 1}$
$\frac{9}{25} = \frac{m}{m + 1}$
$9 m + 9 = 25 m$
$16 m = 9$
$m = \frac{9}{16}$

43​

34​

916​

169​

3=2πkm​​ 5=2πkm+1​​ 53​=m+1m​​ 259​=m+1m​ 9m+9=25m 16m=9 m=169​

$\frac{3}{4}$

$\frac{4}{3}$

$\frac{16}{9}$

$\frac{9}{16}$

$3 = 2 π \frac{m}{k}$
$5 = 2 π \frac{m + 1}{k}$
$\frac{3}{5} = \frac{m}{m + 1}$
$\frac{9}{25} = \frac{m}{m + 1}$
$9 m + 9 = 25 m$
$16 m = 9$
$m = \frac{9}{16}$