Elastic Strings | Hooke's Law | A Level Maths Revision Notes

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1 Summary

1.1 Hooke’s Law

1.2 Example #1

1.3 Elastic Potential Energy

1.4 Example #2

1.4.1 Reference

Summary

An elastic string is one that can be stretched and it will return back to its original length.
Hooke’s Law basically states that within the elastic limits, extension is proportional to the tensile force. It is defined as: $T\quad =\quad \frac { \lambda e }{ l }$ where $\lambda$ is the modulus of the string, T is the tensile force, e is the extension and l is the strings natural length.
Extension can be calculated through the formula: $e\quad =\quad New\quad Length\quad after\quad stretching\quad -\quad Natural\quad Length\quad of\quad the\quad String$
A stretched elastic string spring possess elastic potential energy which is equal to the amount of work done to provide the extension. It is defined as:
$Elastic\quad Potential\quad Energy\quad (E.P.E)\quad =\quad \frac { \lambda { e }^{ 2 } }{ 2l }$

An elastic string is one that can be stretched and it will return back to its original length. When the string is stretched it is called ”taut”. We apply two forces at each end to stretch an elastic string. These forces act outwards and they are equal to the tension in the string.

Hooke’s Law

When some tensile force is applied on an elastic string it produces some extension in the string. Through experimentation, it is found that within the elastic limits, the extension is proportional to the tensile force. This law is known as Hooke’s Law.

Let’s say l is the natural length of the elastic string, and e is the extension produced by a tensile force T. Then Hooke’s law is defined through a formula as:

$T\quad =\quad \frac { \lambda e }{ l }$

Where is $\lambda$ a constant and is known as modulus of the string. The unit of is Newton and its value differs according to the material and geometry of the string.

We can also calculate the extension e of the string if we know the strings natural length and the new total length after stretching.

$e\quad =\quad New\quad Length\quad after\quad stretching\quad -\quad Natural\quad Length\quad of\quad the\quad String$

Also remember that once a string reaches its elastic limit, it will not return back to its original position, therefore it will no longer obey Hooke’s Law.

Example #1

Q. A light elastic string of natural length 50 cm and modulus 40 N has one end fixed. A particle of mass 2 Kg is attached at the other end and hangs in equilibrium. Calculate the extension in the string.

Solution:

Let’s say extension is e meters.
As the particle is hanging vertically and is in equilibrium, we can say:

T = mg where m = 2 kg
T = 2g

Now we apply the Hooke’s Law to find extension e:

$T\quad =\quad \frac { \lambda e }{ l }$

We know $\lambda\quad =\quad40\quad N$ and natural length = 0.5 m

Substitute the values in the above equation:

$2g\quad =\quad \frac { 40e }{ 0.5 }$ as $g\quad =\quad 9.8\quad { m }/{ { s }^{ 2 } }$

$e\quad =\quad \frac { g }{ 40 }$

$e\quad =\quad \frac { 9.8 }{ 40 }$

$e\quad =\quad0.245 \quad m$ Ans

Elastic Potential Energy

A stretched elastic string or a stretched or compressed elastic spring possess elastic potential energy which is equal to the amount of work done to provide the extension or compression.

Let’s say we apply a tensile force T to a string of natural length . The modulus $\lambda$ produces an extension e.

Therefore by Hooke’s Law $T\quad =\quad \frac { \lambda e }{ l }$ and now if we stretch the string a little more say $\delta e$ , than the work done by the force will now be:

$\delta work\quad done\quad =\quad \frac { \lambda e }{ l } \delta e$

Total work done to give extension from ${ e }_{ 1 }\quad to\quad { e }_{ 2 }$ is:

$W\quad =\quad \int _{ e1 }^{ e2 }{ \frac { \lambda e }{ l } le\quad de }$

$W\quad =\quad \left[ \frac { \lambda { e }_{ 2 } }{ l } \quad +\quad \frac { \lambda { e }_{ 1 } }{ l } \right] \frac { ({ e }_{ 2 }\quad -\quad { e }_{ 1 }) }{ 2 }$

$W\quad =\quad \left[ \frac { T_{ 1 } }{ l } \quad +\quad \frac { T_{ 2 } }{ l } \right] \frac { ({ e }_{ 2 }\quad -\quad { e }_{ 1 }) }{ 2 }$

Elastic Potential Energy stored in a string is hence defined through a formula as:

$Elastic\quad Potential\quad Energy\quad (E.P.E)\quad =\quad \frac { \lambda { e }^{ 2 } }{ 2l }$

Example #2

Q. An elastic string has natural length 2 m and modulus 15 N. Calculate the energy stored in the string when the string is stretched to a length of 2.6 m.

Solution:

We know the following data from the question:

l = 2 m, $\lambda\quad =\quad15\quad N$ and stretched length = 2.6 m

We first calculate extension e using the formula:

$e\quad =\quad New\quad Length\quad after\quad stretching\quad -\quad Natural\quad Length\quad of\quad the\quad String$

e = 2.6 – 2

e = 0.6 m

Use the above formula of elastic potential energy and substitute these values.

$Elastic\quad Potential\quad Energy\quad (E.P.E)\quad =\quad \frac { \lambda { e }^{ 2 } }{ 2l }$

$=\quad \frac { 15{ (0.6) }^{ 2 } }{ 2(2) }$

$=\quad 1.35\quad Joules$ Ans