Frequency of an oscillating mass with springs in series

In summary: A + B)*C = A*C + B*CSo you will have two terms in the denominator with a common factor that is the product of two other terms.
  • #1
laurette1029
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Homework Statement



A block on a frictionless table is connected as shown in the figure to two springs having spring constants k1 and k2.

Find an expression for the block's oscillation frequency f in terms of the frequencies f1 and f2 at which it would oscillate if attached to spring 1 or spring 2 alone.
Give your answer in terms of f1 and f2.


Homework Equations



f=(1/2pi)*sqrt(k/m)

The Attempt at a Solution



If the object was attached to spring 1 alone, the frequecy would be:

f1=(1/2pi)*sqrt(k1/m)

If the object was attached to spring 2 alone, the frequecy would be:

f2=(1/2pi)*sqrt(k2/m)

Since the two springs are in a serie, the constant k of the system of spring is :

1/k1+1/k2=1/k which means k=k1k2/k1+k2

Then the frequecy of the system of springs should be f=(1/2pi)*sqrt(k1k2/m(k1+k2))

This is where I get stuck, I don't know how to express f in terms of f1 and f2.
 

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  • #2
laurette1029 said:

Homework Statement



A block on a frictionless table is connected as shown in the figure to two springs having spring constants k1 and k2.

Find an expression for the block's oscillation frequency f in terms of the frequencies f1 and f2 at which it would oscillate if attached to spring 1 or spring 2 alone.
Give your answer in terms of f1 and f2.


Homework Equations



f=(1/2pi)*sqrt(k/m)

The Attempt at a Solution



If the object was attached to spring 1 alone, the frequecy would be:

f1=(1/2pi)*sqrt(k1/m)

If the object was attached to spring 2 alone, the frequecy would be:

f2=(1/2pi)*sqrt(k2/m)

Since the two springs are in a serie, the constant k of the system of spring is :

1/k1+1/k2=1/k which means k=k1k2/k1+k2

Then the frequecy of the system of springs should be f=(1/2pi)*sqrt(k1k2/m(k1+k2))

This is where I get stuck, I don't know how to express f in terms of f1 and f2.


Perhaps you need to use proportion to get rid of a lot of the clutter.

f1=(1/2pi)*sqrt(k1/m) = (1/2pi)*sqrt(1/m)*sqrt(k1) = A*k1
f2=(1/2pi)*sqrt(k2/m) = (1/2pi)*sqrt(1/m)*sqrt(k2) = A*k2

I replaced all those constant/equal bits with the symbol A. The was no significance in me choosing A, I could have used any letter/symbol - except the ones already used as that would be confusing [so m is out of the question]

When you do a similar thing to the combined spring situation you may find something.
 

Related to Frequency of an oscillating mass with springs in series

1. What is the formula for calculating the frequency of an oscillating mass with springs in series?

The formula for calculating the frequency of an oscillating mass with springs in series is:
f = 1/(2π) * √(k/m)
Where f is the frequency, k is the combined spring constant, and m is the mass of the object.

2. How does the mass of the object affect the frequency of the oscillation?

The mass of the object has an inverse relationship with the frequency of the oscillation. This means that as the mass increases, the frequency decreases, and vice versa. This is because a larger mass requires more force to move it, resulting in a slower oscillation.

3. Does the stiffness of the springs affect the frequency of the oscillation?

Yes, the stiffness of the springs, represented by the spring constant (k), directly affects the frequency of the oscillation. A higher spring constant will result in a higher frequency, while a lower spring constant will result in a lower frequency.

4. How does the number of springs in series affect the frequency of the oscillation?

The number of springs in series does not affect the frequency of the oscillation. This is because the combined spring constant (k) takes into account the stiffness of all the springs in the system, regardless of the number of springs.

5. Can the frequency of an oscillating mass with springs in series be altered?

Yes, the frequency of an oscillating mass with springs in series can be altered by changing the mass of the object or the stiffness of the springs. Additionally, external forces such as damping or additional masses can also affect the frequency of the oscillation.

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