Hooke’s Law Equation Experiment - Definition, FAQs

Over the ages, many innovators have worked hard to create a variety of devices. They're all predicated on a thorough comprehension of the mechanics. Before using, operators need to have a basic understanding of mechanics. One of these is Hooke's law.
The most commonly employed device in this law is the spring. This aids in the definition of elasticity, torsion, and force laws. For a play, all of these characteristics come together. Hooke's law is applicable within the spring's valid elastic limit. The only attribute that allows the spring to remain in a limited space is its elasticity. The spring will lose its property if it breaks.

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This Story also Contains

1. Hooke’s Law: Definition
2. Solved Examples Based on Hooke's law
3. Summary

Hooke’s Law: Definition

Hooke's Law is regarded as one of the finest physics principles. This law states that the strain is proportional to stress applied to the material within the elastic limit.

In the 17th century, British physicist Robert Hooke developed. He established a link between the numerous forces applied to a spring and its elasticity.

Hooke’s law equation:

F_S=-kx

Hooke's law is applicable within the spring's valid elastic limit. The only attribute that allows the spring to remain in a limited space is its elasticity. The spring will lose its property if it breaks.

Hooke's Law, on the other hand, only works in a narrow context. This is analogous to the most fundamental law of mechanics. This is due to the fact that no material can be crushed or stretched beyond a specific minimum or maximum size.

State Hooke's Law of Elasticity

There is no way to permanently distort or modify the state of the spring. Hooke's law is only relevant when a little amount of force or deformation is involved. Consider the fact that many materials will diverge dramatically from Hooke's law. This is due to their extremely pliable boundaries.

Some general types of physics can relate Hooke's Law to Newton's rules of static equilibrium since they are mutually compatible. When strain and stress are evaluated simultaneously, the accurate relationship between strain and stress for complicated objects may be traced.

This relationship is entirely dependent on the intrinsic qualities of materials. Consider the case of a homogeneous rod with a uniform cross-section. During the stretching, this rod will function as a basic spring.

The stiffness (k) of the rod is related to the area of the rod's cross-section. Too, according to the law of elasticity, it is inversely proportional to its length.

Hooke’s Law Derivation

In mathematical terms, the following equation can be used to represent Hooke's law:

Hooke’s law equation is:

F_S=-kx

F is the applied force, and it is equal to a constant in this equation. K is a constant that equals k times the displacement or change in length of an item symbolized by x.

Where,

F = Force applied

k = Displacement constant

x = the object's length

The amount of k is affected by the type of elastic material, its size, and its shape. When a significant amount of force is applied, the elastic material deforms several times more than the amount predicted by Hooke's Law. The material, on the other hand, remains elastic and returns to its original size when the force is released, and it also retains its shape when the force is removed. Hooke's Law equation can sometimes be written as follows:

Spring restoring force = - Spring constant x spring displacement

F_s = -kx

Where,

F = spring restoring force (Newtons, N)

Spring constant (N/m) = k

x = the spring's displacement (m)

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Solved Examples Based on Hooke's law

Example 1: Hooke's law essentially defines

1) stress

2) strain

3) yield point

4) elastic limit

Solution

Hooke's Law

stress ∝ strain

stress strain =E

wherein

E= modulus of elasticity

Hence, the answer is the option (4).

Example 2: A uniformly tapering conical wire is made from a material of Young’s modulus Y and has a normal, unextended length L. The radii, at the upper and lower ends of this conical wire, have values R and 3R, respectively. The upper end of the wire is fixed to a rigid support and a mass M is suspended from its lower end. The equilibrium extended length, of this wire, would equal

1) L(1+29MgπYR2)
2) L(1+13MgπYR2)
3) L(1+19MgπYR2)
4) L(1+23MgπYR2)

Solution:

Young Modulus

The ratio of normal stress to longitudinal strain

it denoted by Y
Y= Normal stress longitudnal strain

wherein

Y=F/AΔl/L
F - applied force
A - Area
Δl - Change in length
I- original length

r−Rx=3R−RLr=R(1+2xL)Y=mgπR2dLdx⇒dL=mgπR2dx(1+2xL)2ΔL=mgYπR2∫0Ldx(1+2xL)2⇒mgL(1+2xL)2L′=L+ΔL=L+mgL(1+2xL)2 Now, L′=L(1+13mgπR2Y)

Hence, the answer is the option (2).

Example 3: A wire elongates by l mm when a load W is hung from it. If the wire goes over a pulley and two weights W each are hung at the two ends, the elongation of the wire will be (in mm )

1) l/2
2) l
3) 2 L
4) zero

Solution:

Young Modulus

The ratio of normal stress to longitudinal strain

it denoted by Y
Y= Normal stress longitudnal strain

wherein

Y=F/AΔl/L
F - applied force
A - Area
Δl - Change in length
1- original length

At equilibrium T=w

y=W/Al/L
Now in case (ii) T=w
let new elongation is Δl

⇒y=w/AΔl/L
So Δl=l

i.e. Elongation is the same.

Hence, the answer is the option (2).

Example 4: A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1%. What is the fundamental frequency (in Hz ) of steel if the density and elasticity of steel are 7.7×103 kg/m3 and 2.2×1011 N/m2 respectively?

1) 770

2) 188.5

3) 178.2

4) 200.5

Solution:

Young Modulus

The ratio of normal stress to longitudinal strain

it denoted by Y
Y= Normal stress longitudnal strain Y=F/AΔl/L F- applied force A - Area Δl - Change in length 1- original length Fundamental frequency v=12l⋅Tμ=12lTρA12l⋅ stress ρ=12×1.5Y× strain ρ13⋅2.2×1011×10−27.7×103=178.2 Hz
Hence, the answer is the option (3).

Example 5: Young's moduli of two wires A and B are in the ratio 7:4. Wire A is 2 m long and has a radius of R. Wire B is 1.5 m long and has a radius 2 mm. If the two wires stretch by the same length for a given load, then the value of R (in mm) is close to:

1) 1.5

2) 1.9

3) 1.7

4) 1.3

Solution:

Y=Flπr2ΔlYAYB=FlAπrA2ΔlA×πrB2ΔlBFlB74=2(rA)2×221.5⇒rA2=4×2×221.5×7⇒rA=1.7 mm

Hence, the answer is the option (3).

Summary

Hooke's Law explains the relationship between the force applied to an elastic material and its deformation, stating that stress is proportional to strain within the elastic limit. It is primarily demonstrated using springs, with elasticity being a key property. The law breaks down when materials exceed their elastic limits. Several examples and mathematical derivations illustrate its application in various physics problems, such as calculating elongation, tension, and fundamental frequencies in wires