Question

For given spring mass system if the time period of small oscillations of block about its mean position

Answer 1

$\mathrm{T}= \pi \sqrt{m / k}$

Explanation:

When $k_{2}$ moves by x

and $k_{1}$  moves by 2x.

2T=  $k_{1}$(2x)

and T =  $k_{2}$ so, both are in parallel.

Keff = $k_{1}$ + $k_{2}$

So,

$\mathrm{T}=2 \pi \sqrt{m / 4k}$

$\mathrm{T}= \pi \sqrt{m / k}$

To learn more

i) A spring stores 5 J of energy when stretched by 25 cm. It is kept vertical with the lower end fixed. A block fastened to its other end is made to undergo small oscillations. If the block makes 5 oscillations each second, what is the mass of the block?

https://brainly.in/question/7650745

ii)The potential energy of a particle of mass 1kg in motion along the x axis is given by U =4(1-cos 2x)J , where x is in metres . The period of small oscillations (in s) is

https://brainly.in/question/6575071

Answer 2

Complete question:

For the given spring mass system, find the time period of small oscillations of block about its mean position. Assume ideal conditions. The system is in vertical plane.

Answer:

The time period of small oscillations of block is 2π √(m/k)

Explanation:

If the spring k₂ moves x distances then spring k₁ moves by 2x distance.

The time period of the spring k₁ is given as:

2T= k₁(2x)

The time period of the spring k₂ is given as:

T = k₂x

Since, both the springs are parallel to each other, then,

Keff = k₁ + k₂

Thus, time period of the block about its mean position is:

∴ T = 2π √(m/k)