If a spring is cut into 9 equal pieces then they r connected in series with the same mass.what will be it's frequency relative to the original frequency?my text book says that the answer is 9f.but Time period of spring does not depend on length.right?

• "If a spring is cut into 9 equal pieces then they r connected in series with the same mass.what will be it's frequency relative to the original frequency?my text book says that the answer is 9f.but Time period of spring does not depend on length.right?"

There is clearly a typo in here. If you re-connect the same 9 equal pieces in series , you have just glued the spring back together, and the spring will behave as it did before

I think you mean that the springs are then used in parallel . Which means digging deeper into the underlying ideas.

• Spring Constants

The spring constant is a idealised constant for a spring of a given length . The more fundamental relationship, which actually applies to a wire in simple extension but drives the whole inter-molecular elasticity idea, is between stress, #sigma#, and strain, #varepsilon#, and involves Young's Modulus, #Y#. It is not only a more complicated but more intuitive too.

#Y = (sigma)/(varepsilon) = (F/A)/((Delta L)/(L_o)) = " const" qquad triangle#

• #F# is the applied force, #A# is the spring's cross-sectional area, and #Delta L# is the increase or decrease in length of that particular spring of unextended length #L_o#. Common sense: A thicker wire will be more opposed to any extension, any extension is proportional to the original length of the wire.

We can manipulate #triangle# to get this result, which looks more like Hooke's Law:

• #F = underbrace( (Y A )/ L_o)_("k") * underbrace(Delta L)_("x") #

Excluding the constants:

• #k propto 1/L_o#

So, if we cut a spring into 9 pieces, #L_o' = 1/9 L_o#, and:

• #(k')/(k) = (L_o')/(L_o) = 9 implies k' = 9 k#

The spring is suddenly 9-times more "springier" for Hooke's Law , even though it is the same spring, just cut up.

• Combining the springs in parallel

When the 9 springs are in parallel, each spring will experience, under Hooke's Law a force, #F'#, that is:

• #F' = F/9#

Putting it together :

• #F' = k' x#

• #F/9 = 9 k x qquad implies F = bb(81 \ k) \ x color(blue)( implies k' = 81 \ k)#

You should know that:

• #omega = sqrt(k/m)#

• #omega' = sqrt((k')/(m')) = sqrt((81k)/m) = bb9 \ omega#

Here's another approach. It turns out that the frequency stays the same.

From what you describe, the springs are all in-phase:

(pretend there are 9 springs here.)

Suppose you had #9# small springs with force constant #k_i = 9k#, since you chopped up one spring of force constant #k#. We pretend to ignore the mass of the springs.

If the force is the same on all #9# springs, then from Hooke's law,

#F_k = -k_1x_1 = . . . = -k_9x_9#

Solving the first two of these equations gives:

#-k_1x_1 = -k_9x_9#

From this,

#x_1 = k_9/k_1x_9#.

Continuing, we would also get everything in terms of #x_9#:

#x_2 = k_9/k_2x_9#

#x_3 = k_9/k_3x_9#

#vdots#

#x_8 = k_9/k_8x_9#

The effective force for the system would be:

#F_k = -k_(eff)(x_1 + x_2 + . . . + x_9)#

The force for each component of the system must be the same, because it is externally imparted in the same uniform phase, so:

#-k_ix_i = -k_(eff)(x_1 + x_2 + . . . + x_9)#

For the 9th spring component then,

#-k_9x_9 = -k_(eff)(x_1 + x_2 + . . . + x_9)#

#= -k_(eff)(k_9/k_1x_9 + k_9/k_2x_9 + . . . + x_9)#

Dividing through by #k_9x_9#:

#-1 = -k_(eff)(1 / k_1 + 1/k_2 + . . . + 1/k_9)#

The effective force constant of the system is then:

#k_(eff) = (sum_(i=1)^(9) 1/k_i)^(-1)#

The frequency of oscillation is then:

#omega = sqrt(k_(eff)/m)#

#= sqrt((sum_(i=1)^(9) 1/k_i)^(-1)/m)#

Now, each force constant for a given spring component is the same (since the spring is the same material), so #k_1 = . . . = k_9#. Therefore:

#omega = sqrt((9//k_i)^(-1)/m)#

#= sqrt(k_i/(9m)) = 1/3sqrt(k_i/m)#

Lastly, each force constant is #9# times the original, total force constant, meaning #k_i = 9k#, since #1//9# the length is #9# times stiff.

As a result,

#omega = sqrt((9k)/(9m)) = sqrt(k/m) = omega#.

The frequency stays the same.