JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Elasticity & SHM

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

\frac{T _{b}}{T _{a}} = x

, where value of

x

is ____ (Round off to the Nearest Integer)

Answer: 2

Solution

A single spring of constant

k

carrying a mass

M

oscillates with

T_{a} = 2 π \frac{M}{k}

Two such springs in series stretch twice as much under the same tension, so the combination has constant

k /2

and

T_{b} = 2 π \frac{2 M}{k}

Dividing the two periods,

\frac{T _{b}}{T _{a}} = 2

Comparing with

x

gives

x = 2

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Integer answer