Question

Two identical springs of spring constant k are attached to a block of mass m and to fixed supports as shown in the figure. The time period of oscillation is

• A.
• B. $2 \pi \sqrt { \frac { \mathrm {~m} } { 2 \mathrm { k } } }$
• C. $2 \pi \sqrt { \frac { 2 \mathrm {~m} } { \mathrm { k } } }$
• D. $\pi \sqrt { \frac { \mathrm { m } } { 2 \mathrm { k } } }$

Answer 1

Answer: (B)

$2 \pi \sqrt { \frac { \mathrm {~m} } { 2 \mathrm { k } } }$

Answer 2

Let the mass m be displace by a small distance x to the right from its mean position as shown in figure (2). due to it the spring on the left side gets stretched by a length x while that on the right side gets compressed by the same length.

The forces acting on the mass are

$\mathrm { F } _ { 1 } = - \mathrm { kx }$ towards left hand side

$\mathrm { F } _ { 2 } = - \mathrm { kx }$ towards left hand side

The net force acting on the mass is

$\mathrm { F } = \mathrm { F } _ { 1 } + \mathrm { F } _ { 2 } = - 2 \mathrm { kx }$

Here, and –ve sign shows that force is towards the mean position therefore the motion executed by the particle is simple harmonic.

Its acceleration is

$\mathrm { a } = \frac { \mathrm { F } } { \mathrm { m } } = - \frac { 2 \mathrm { kx } } { \mathrm { m } }$ …… (i)

The standard equation of SHM is

$a = - \omega ^ { 2 } x$ ……. (ii)

Comparing (i) and (ii) we get

Time period $\mathrm { T } = \frac { 2 \pi } { \omega } = 2 \pi \sqrt { \frac { \mathrm {~m} } { 2 \mathrm { k } } }$