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An object of mass m is attached by two stretched springs (stiffnessKs and relaxed lengthL0 ) to rigid walls, as shown in figure 4.60. The springs are initially stretched by an amount (L-L0). When the object is displaced to the right and released, it oscillates horizontally. Starting from the momentum principle, find a function of the displacement xof the object and the timet describes the oscillatory motion.

(a) What is the period of the motion?

(b) If L0 were shorter (so the springs are initially stretched more), would the period be larger, smaller, or the same?

Short Answer

Expert verified
  1. The time period is2m2ks .
  2. As the expression of time period is independent of length, so on increasing or decreasing the length L0, time period will not change.

Step by step solution

01

Understanding the concept

The expression for the momentum is given by,

p=mv 鈥︹. (1)

Here pis the momentum, mis the mass, vis the velocity.

The rate of change of momentum is known as force and it is expressed as,

F=dpdt鈥︹ (2)

HereF is the force.

02

Calculate the period of the motion.

(a)

The net force is equal to the 2ks,

Substitute mv for pand 2ksfor Finto the equation (2),

2ks=d(mv)dt=md2xdt2

From the above expression,

d2xdt2=2ksmx 鈥︹ (3)

Here x is the displacement.

Equation (3) is in the form d2xdt2=2x

Here is the angular velocity,

2=2ksm=2ksm

The expression for the time period is given by,

T=2

Substitute 2ksm for in the above equation,

T=2m2ks

Therefore the time period is2m2ks .

03

Dependency of time period on length.

(b)

As the expression of time period is independent of length, so on increasing or decreasing the length L0, time period will not change.

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