Question 194: Oscillations

Question

Consider two identical springs each of spring constant k and negligible mass compared to the mass M as shown. Fig. 1 shows one of them and Fig. 2 shows their series combination. The ratios of time period of oscillation of the two SHM is T
b
/T
a
= $$\sqrt x $$, where value of x is ___________. (Round off to the Nearest Integer)

Accepted Answer

$${T_a} = 2\pi \sqrt {{M \over k}} $$
$${K_{e{q_{series}}}} = {{{k_1}{k_2}} \over {{k_1} + {k_2}}} = {k \over 2}$$
$${T_b} = 2\pi \sqrt {{M \over {k/2}}}  = 2\pi \sqrt {{{2M} \over k}} $$
$${T_b} = \sqrt 2 {T_a}$$
$${{{T_b}} \over {{T_a}}} = \sqrt 2 $$
$$ 	herefore $$ x = 2

Oscillations

Integer / Numeric Answer Type