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Q. 75

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Physics for Scientists and Engineers: A Strategic Approach with Modern Physics
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Short Answer

A block on a frictionless table is connected as shown in FIGURE P15.75 to two springs having spring constants and . Find an expression for the block’s oscillation frequency in terms of the frequencies and at which it would oscillate if attached to spring or spring alone.

The expression for the block’s oscillation frequency

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Step by Step Solution

Step 1 Introduction.

Newton's Second Law states that the net force acting on a mass body is proportional to the acceleration of the body.

According to Hooke's Law, any thing that causes a spring to extend or compress exerts an elastic force on the object. The -component of the force the spring exerts on the item if the object strains the spring in the -direction is:

Step 2 The Principles.

where is the spring constant in newtons per metre, which is a measure of the spring's (or any elastic object's) stiffness, and is the stretch/compression distance (not the total length of the object). The spring's elastic force on the object is in the opposite direction of how it was stretched (or compressed), hence the negative sing in front of . The spring is then acted upon by the object:

Step 3  Principles.

In basic harmonic motion, the frequency of an oscillator is given by

It's worth noting that is solely affected by the mass and the force constant , not the amplitude.

Step 4 The Given data and Required data.

Given data:

The following are the spring constants for the two springs: and

Figure depicts a mass block on a frictionless table attached to two springs.

If the block is only coupled to springs or , its oscillation frequencies are: and

Required data:

The oscillation frequency of the block must be determined.

Step 5 Solution.

Because the mass is exclusively in touch with spring , it exerts a force on it . Furthermore, because spring exerts a force according to Equation , the net force on spring must be zero.

Using Hooke's law from Equation , can be found:

where is the spring displacement.

Step 6

Spring is similarly affected by the wall, which exerts a force of . In addition, spring exerts a force on on spring in the opposite direction as . Because spring is likewise supposed to be massless, the net force exerted on it must be zero, according to Newton's second law. So

Using Hooke's law from Equation , we can find :

where is the spring displacement.

Step 7

The force on exerted by spring on spring 1 is equal in size to the force exerted by spring on spring according to Newton's third law. So

Substitute from Equation and from Equation (4):

Step 8

The force exerted by the mass on spring is also equivalent in size to the force exerted by spring on the mass, according to Newton's third law.

Since spring is the only spring exerting a force on the mass, we utilised instead of .

Step 9

The length change of the complete system of two springs will be equal to the length change of each spring individually:

Substitute and in Equation :

When we solve for ,we get:

where is the system's effective spring constant.

Step 10

Equation gives the frequency of oscillation of the mass if it is only coupled to spring :

Rearrange the equations and solve for :

Step 11

Similarly, if the mass is just attached to spring , the frequency of oscillation is:

Rearrange the equations and solve for :

Step 12

Equation can also be used to calculate the frequency of oscillation of the mass and two springs in terms of the effective spring constant of the system:

Substitute and from Equations and :

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