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

Options

A. `2pisqrt((m)/(k))`
B. `2pisqrt((m)/(k))`
C. `2pisqrt((2m)/(k))`
D. `pisqrt((m)/(2k))`

Answer

B

Explanation

Let the mass m be displaced by a small distance x to the right from its mean position as shown in figure (b). Due to it’s the spring on the left side gets stretched by a length x while that on the right side gets compressed by the sae length. the forces acting on the mass are the same length. the forces acting on the mass are
`F_(1)=-kx` towards left hand side
`F_(2)=-kx` towards left hand side
The net force acting on the mass is, `F=F_(1)=-2kx`
Here, `F propx and -ve` sign shows that force is towards the mean position, therefore the motion executed by the particle is simple harmonic.
Its acceleration is
`a=(F)/(m)=-(2kx)/(m)` . . . (i)
the standard equation of SHM is
`a=-omega^(2)x` . . . (ii)
Comparing (i) and (ii), we get
`omega^(2)=(2k)/(m)` or `omega=sqrt((2k)/(m))`
time period, `T=(2pi)/(omega)=2pisqrt((m)/(2k))`