Question

In the system shown in the figure, the string, springs and pulley are weightless. The force constants of the two springs are k1=k and k2=2k. Block of mass M is pulled vertically down from its equilibrium position and released. Calculate the angular frequency of oscillation.

[Assume the top surface of the block (represented by line AB) always remains horizontal]

[Assume the top surface of the block (represented by line AB) always remains horizontal]

- ω=√4k3M
- ω=√8k3M
- ω=√k3M
- ω=√2k3M

Solution

The correct option is **B** ω=√8k3M

Let x0 be the elongation in spring of force constant k2 at equilibrium.

From the data given in the question, it is clear that the elongation in the other spring must be 2x0, so that tension is the same in both the springs.

[since both the springs are joined by the same piece of string, spring forces in each must be equal to tension in the rope]

For equilibrium of block,

2kx0+k(2x0)=Mg

⇒4kx0=Mg ……(1)

Let the block be displaced downward by x, causing the two springs to stretch further by x1 and x2 respectively.

Again, k1x1=k2x2⇒x1=2x2

and x1+x2=2x [from string constraint]

∴x2=2x3 and x1=4x3

For the block displaced by distance x from equilibrium/mean position,

Md2xdt2=Mg−k1(x1+2x0)−k2(x2+x0)

⇒Md2xdt2=Mg−k1(x1+2x0)−k2(x2+x0)

⇒Md2xdt2=Mg−kx1−2kx2−(2k+2k)x0

⇒Md2xdt2=−4kx3−4kx3 [using (i)]

⇒d2xdt2=−8k3Mx

Comparing this with a=−ω2x,

we get, ω=√8k3M

Thus, option (b) is the correct answer.

Let x0 be the elongation in spring of force constant k2 at equilibrium.

From the data given in the question, it is clear that the elongation in the other spring must be 2x0, so that tension is the same in both the springs.

[since both the springs are joined by the same piece of string, spring forces in each must be equal to tension in the rope]

For equilibrium of block,

2kx0+k(2x0)=Mg

⇒4kx0=Mg ……(1)

Let the block be displaced downward by x, causing the two springs to stretch further by x1 and x2 respectively.

Again, k1x1=k2x2⇒x1=2x2

and x1+x2=2x [from string constraint]

∴x2=2x3 and x1=4x3

For the block displaced by distance x from equilibrium/mean position,

Md2xdt2=Mg−k1(x1+2x0)−k2(x2+x0)

⇒Md2xdt2=Mg−k1(x1+2x0)−k2(x2+x0)

⇒Md2xdt2=Mg−kx1−2kx2−(2k+2k)x0

⇒Md2xdt2=−4kx3−4kx3 [using (i)]

⇒d2xdt2=−8k3Mx

Comparing this with a=−ω2x,

we get, ω=√8k3M

Thus, option (b) is the correct answer.

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