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?
2 Answers
Your book looks right :)
Explanation:
- "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,
#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
#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,
#(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' = 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
If the force is the same on all
#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_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
#-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
#omega = sqrt((9//k_i)^(-1)/m)#
#= sqrt(k_i/(9m)) = 1/3sqrt(k_i/m)#
Lastly, each force constant is
As a result,
#omega = sqrt((9k)/(9m)) = sqrt(k/m) = omega# .
The frequency stays the same.
