Chapter9 Mechanical Properties Of Solids
Chapter Nine
MECHANICAL PROPERTIES OF SOLIDS
MCQ I
9.1 Modulus of rigidity of ideal liquids is
(a) infinity.
(b) zero.
(c) unity.
(d) some finite small non-zero constant value.
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Answer: (b)9.2 The maximum load a wire can withstand without breaking, when its length is reduced to half of its original length, will
(a) be double.
(b) be half.
(c) be four times.
(d) remain same.
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Answer: (d)9.3 The temperature of a wire is doubled. The Young’s modulus of elasticity
(a) will also double.
(b) will become four times.
(c) will remain same.
(d) will decrease.
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Answer: (d)9.4 A spring is stretched by applying a load to its free end. The strain produced in the spring is
(a) volumetric.
(b) shear.
(c) longitudinal and shear.
(d) longitudinal.
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Answer: (c)9.5 A rigid bar of mass $M$ is supported symmetrically by three wires each of length $l$. Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to
(a) $Y_{\text{copper }} / Y_{\text{iron }}$
(b) $\sqrt{\frac{Y_{\text {iron }}}{Y_{\text {copper }}}}$
(c) $\frac{Y_{\text {iron }}^{2}}{Y_{\text {copper }}^{2}}$
(d) $\frac{Y_{\text {iron }}}{Y_{\text {copper }}}$.
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Answer: (b)9.6 A mild steel wire of length $2 L$ and cross-sectional area $A$ is stretched, well within elastic limit, horizontally between two pillars (Fig. 9.1). A mass $m$ is suspended from the mid point of the wire. Strain in the wire is
Fig. 9.1
(a) $\frac{x^{2}}{2 L^{2}}$
(b) $\frac{x}{L}$
(c) $\frac{x^{2}}{L}$
(d) $\frac{x^{2}}{2 L}$.
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Answer: (a)9.7 A rectangular frame is to be suspended symmetrically by two strings of equal length on two supports (Fig. 9.2). It can be done in one of the following three ways;
(a)
(b)
(c)
Fig. 9.2
The tension in the strings will be
(a) the same in all cases.
(b) least in (a).
(c) least in (b).
(d) least in (c).
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Answer: (c)9.8 Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both the rods are fixed rigidly at one end to the roof. A mass $M$ is attached to each of the free ends at the centre of the rods.
(a) Both the rods will elongate but there shall be no perceptible change in shape.
(b) The steel rod will elongate and change shape but the rubber rod will only elongate.
(c) The steel rod will elongate without any perceptible change in shape, but the rubber rod will elongate and the shape of the bottom edge will change to an ellipse.
(d) The steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre.
MCQ II
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Answer: $(\mathrm{~d})$9.9 The stress-strain graphs for two materials are shown in Fig.9.3 (assume same scale).
Fig. 9.3
(a) Material (ii) is more elastic than material (i) and hence material (ii) is more brittle.
(b) Material (i) and (ii) have the same elasticity and the same brittleness.
(c) Material (ii) is elastic over a larger region of strain as compared to (i).
(d) Material (ii) is more brittle than material (i).
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Answer: (c), (d)9.10 A wire is suspended from the ceiling and stretched under the action of a weight $F$ suspended from its other end. The force exerted by the ceiling on it is equal and opposite to the weight.
(a) Tensile stress at any cross section $A$ of the wire is $F / A$.
(b) Tensile stress at any cross section is zero.
(c) Tensile stress at any cross section $A$ of the wire is $2 F / A$.
(d) Tension at any cross section $A$ of the wire is $F$.
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Answer: (a), (d)9.11 A rod of length $l$ and negligible mass is suspended at its two ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths (Fig. 9.4). The cross-sectional areas of wires A and B are $1.0 \mathrm{~mm}^{2}$ and $2.0 \mathrm{~mm}^{2}$, respectively.
$$ \left(Y_{A l}=70 \times 10^{9} \mathrm{Nm}^{-2} \text { and } Y_{\text {steel }}=200 \times 10^{9} \mathrm{Nm}^{-2}\right) $$
(a) Mass $m$ should be suspended close to wire $A$ to have equal stresses in both the wires.
(b) Mass $m$ should be suspended close to B to have equal stresses in both the wires.
(c) Mass $m$ should be suspended at the middle of the wires to have equal stresses in both the wires.
(d) Mass $m$ should be suspended close to wire A to have equal strain in both wires.
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Answer: (b), (d)9.12 For an ideal liquid
(a) the bulk modulus is infinite.
(b) the bulk modulus is zero.
(c) the shear modulus is infinite.
(d) the shear modulus is zero.
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Answer: (a), (d)9.13 A copper and a steel wire of the same diameter are connected end to end. A deforming force $F$ is applied to this composite wire which causes a total elongation of $1 \mathrm{~cm}$. The two wires will have
(a) the same stress.
(b) different stress.
(c) the same strain.
(d) different strain.
VSA
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Answer: (a), (d)9.14 The Young’s modulus for steel is much more than that for rubber. For the same longitudinal strain, which one will have greater tensile stress?
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Answer: Steel9.15 Is stress a vector quantity?
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Answer: No9.16 Identical springs of steel and copper are equally stretched. On which, more work will have to be done?
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Answer: Copper9.17 What is the Young’s modulus for a perfect rigid body?
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Answer: Infinite9.18 What is the Bulk modulus for a perfect rigid body?
SA
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Answer: Infinite
Exemplar Problems-Physics
9.19 A wire of length $L$ and radius $r$ is clamped rigidly at one end. When the other end of the wire is pulled by a force $f$, its length increases by $l$. Another wire of the same material of length $2 L$ and radius $2 r$, is pulled by a force $2 f$. Find the increase in length of this wire.
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Answer: Let $Y$ be the Young’s modulus of the material. Then
$Y=\frac{f / \pi r^{2}}{l / L}$
Let the increase in length of the second wire be $l^{\prime}$. Then
$\frac{\frac{2 f}{4 \pi r^{2}}}{l^{\prime} / 2 L}=Y$
Or, $l^{\prime}=\frac{1}{Y} \frac{2 f}{4 \pi r^{2}} 2 L=\frac{l}{L} \frac{\pi r^{2}}{f} \times \frac{2 f}{4 \pi r^{2}} 2 L=l$
9.20 A steel $\operatorname{rod}\left(Y=2.0 \times 10^{11} \mathrm{Nm}^{-2}\right.$; and $\left.\alpha=10^{-50} \mathrm{C}^{-1}\right)$ of length $1 \mathrm{~m}$ and area of cross-section $1 \mathrm{~cm}^{2}$ is heated from $0^{\circ} \mathrm{C}$ to $200^{\circ} \mathrm{C}$, without being allowed to extend or bend. What is the tension produced in the rod?
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Answer: Because of the increase in temperature the increase in length per unit length of the rod is
$\frac{\Delta l}{l_{0}}=\alpha \Delta T=10^{-5} \times 2 \times 10^{-2}=2 \times 10^{-3}$
Let the compressive tension on the rod be $T$ and the cross sectional area be $a$, then
$\frac{T / a}{\Delta l / l_{0}}=Y$
$\therefore T=Y \frac{\Delta l}{l_{O}} \times a=2 \times 10^{11} \times 2 \times 10^{-3} \times 10^{-4}$
$=4 \times 10^{4} \mathrm{~N}$
9.21 To what depth must a rubber ball be taken in deep sea so that its volume is decreased by $0.1 %$. (The bulk modulus of rubber is $9.8 \times 10^{8} \mathrm{~N} \mathrm{~m}^{-2}$; and the density of sea water is $10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$.)
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Answer: Let the depth be $h$, then the pressure is
$$ P=\rho g h=10^{3} \times 9.8 \times h $$
Now $\left|\frac{P}{\Delta V / V}\right|=B$
$\therefore P=B \frac{\Delta V}{V}=9.8 \times 10^{8} \times 0.1 \times 10^{-2}$
$\therefore h=\frac{9.8 \times 10^{8} \times 0.1 \times 10^{-2}}{9.8 \times 10^{3}}=10^{2} \mathrm{~m}$
9.22 A truck is pulling a car out of a ditch by means of a steel cable that is $9.1 \mathrm{~m}$ long and has a radius of $5 \mathrm{~mm}$. When the car just begins to move, the tension in the cable is 800 N. How much has the cable stretched? (Young’s modulus for steel is $2 \times 10^{11} \mathrm{Nm}^{-2}$.)
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Answer: Let the increase in length be $\Delta l$, then
$$ \begin{gathered} \frac{800}{\left(\pi \times 25 \times 10^{-6}\right) /(\Delta l / 9.1)}=2 \times 10^{11} \ \therefore \Delta l=\frac{9.1 \times 800}{\pi \times 25 \times 10^{-6} \times 2 \times 10^{11}} \mathrm{~m} \ 0.5 \times 10^{-3} \mathrm{~m} \end{gathered} $$
9.23 Two identical solid balls, one of ivory and the other of wet-clay, are dropped from the same height on the floor. Which one will rise to a greater height after striking the floor and why?
LA
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Answer: As the ivory ball is more elastic than the wet-clay ball, it will tend to retain its shape instantaneously after the collision. Hence, there will be a large energy and momentum transfer compared to the wet clay ball. Thus, the ivory ball will rise higher after the collision.9.24 Consider a long steel bar under a tensile stress due to forces $\mathbf{F}$ acting at the edges along the length of the bar (Fig. 9.5). Consider a plane making an angle $\theta$ with the length. What are the tensile and shearing stresses on this plane?
Fig. 9.5
(a) For what angle is the tensile stress a maximum?
(b) For what angle is the shearing stress a maximum?
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Answer: Let the cross sectional area of the bar be $A$. Consider the equilibrium of the plane $a a^{\prime}$. A force $\mathbf{F}$ must be acting on this plane making an angle $\frac{\pi}{2}-\theta$ with the normal ON. Resolving $\mathbf{F}$ into components, along the plane and normal to the plane
$F_{P}=F \cos \theta$
$F_{N}=F \sin \theta$
Let the area of the face $a a^{\prime}$ be $A^{\prime}$, then
$\frac{A}{A^{\prime}}=\sin \theta$
$\therefore A^{\prime}=\frac{A}{\sin \theta}$
The tensile stress $T=\frac{F \sin \theta}{A^{\prime}}=\frac{F}{A} \sin ^{2} \theta$ and the shearing stress $Z=\frac{F \cos \theta}{A^{\prime}}=\frac{F}{A} \cos \theta \sin \theta=\frac{F \sin 2 \theta}{2 A}$. Maximum tensile stress is when $\theta=\pi / 2$ and maximum shearing stress when $2 \theta=\pi / 2$ or $\theta=\pi / 4$.
9.25 (a) A steel wire of mass $\mu$ per unit length with a circular cross section has a radius of $0.1 \mathrm{~cm}$. The wire is of length $10 \mathrm{~m}$ when measured lying horizontal, and hangs from a hook on the wall. A mass of $25 \mathrm{~kg}$ is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains < longitudinal strains, find the extension in the length of the wire. The density of steel is $7860 \mathrm{~kg} \mathrm{~m}^{-3}$ (Young’s modules $\mathrm{Y}=2 \times 10^{11} \mathrm{Nm}^{-2}$ ).
(b) If the yield strength of steel is $2.5 \times 10^{8} \mathrm{Nm}^{-2}$, what is the maximum weight that can be hung at the lower end of the wire?
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Answer: (a) Consider an element $d x$ at a distance $x$ from the load ( $x=0$ ). If $T$ (x) and $T(x+d x)$ are tensions on the two cross sections a distance $\mathrm{d} x$ apart, then
$$ \begin{aligned} & T(x+d x)-T(x)=\mu g d x \text { (where } \mu \text { is the mass/length) } \ & \frac{d T}{d x} d x=\mu g d x \ & \Rightarrow T(x)=\mu g x+C \ & \text { At } x=0, T(0)=M g \Rightarrow C=M g \ & \therefore T(x)=\mu g x+M g \end{aligned} $$
Let the length $\mathrm{d} x$ at $x$ increase by $\mathrm{d} r$, then
$$ \begin{aligned} & \frac{T(x) / A}{\mathrm{~d} r / \mathrm{d} x}=Y \ & \text { or, } \frac{\mathrm{d} r}{\mathrm{~d} x}=\frac{1}{Y A} T(x) \ & \Rightarrow r=\frac{1}{Y A} \int_{0}^{\mathrm{L}}(\mu g x+M g) \mathrm{d} x \ & =\frac{1}{Y A}\left[\frac{\mu g x^{2}}{2}+M g x\right]_{0}^{\mathrm{L}} \ & =\frac{1}{Y A}\left[\frac{m g l}{2}+M g L\right] \end{aligned} $$
( $m$ is the mass of the wire)
$A=\pi \times\left(10^{-3}\right)^{2} \mathrm{~m}^{2}, Y=200 \times 10^{9} \mathrm{Nm}^{-2}$
$m=\pi \times\left(10^{-3}\right)^{2} \times 10 \times 7860 \mathrm{~kg}$
$\therefore r=\frac{1}{2 \times 10^{11} \times \pi \times 10^{-6}}\left[\frac{\pi \times 786 \times 10^{-7} \times 10 \times 10}{2}+25 \times 10 \times 10\right]$
$=\left[196.5 \times 10^{-6}+3.98 \times 10^{-3}\right]-4 \times 10^{-3} \mathrm{~m}$
(b) The maximum tension would be at $x=\mathrm{L}$.
$T=\mu g L+M g=(m+M) g$
The yield force
$=250 \times 10^{6} \times \pi \times\left(10^{-3}\right)^{2}=250 \times \pi N$
At yield
$(m+M) g=250 \times \pi$
$m=\pi \times\left(10^{-3}\right)^{2} \times 10 \times 7860 \quad<M \quad \therefore M g \quad 250 \times \pi$
Hence, $M=\frac{250 \times \pi}{10}=25 \times \pi \quad 75 \mathrm{~kg}$.
9.26 A steel rod of length $2 l$, cross sectional area $A$ and mass $M$ is set rotating in a horizontal plane about an axis passing through the centre. If $Y$ is the Young’s modulus for steel, find the extension in the length of the rod. (Assume the rod is uniform.)
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Answer: Consider an element at $r$ of width $d r$. Let $T(r)$ and $T(r+d r)$ be the tensions at the two edges.
$-T(r+d r)+T(r)=\mu \omega^{2} r d r$ where $\mu$ is the mass/length $-\frac{d T}{d r} d r=\mu \omega^{2} r d r$
At $r=l \quad T=0$
$\Rightarrow C=\frac{\mu \omega^{2} l^{2}}{2}$
$\therefore T(r)=\frac{\mu \omega^{2}}{2}\left(l^{2}-r^{2}\right)$
Let the increase in length of the element $d r$ be $d(\delta)$
$\mathrm{Y}=\frac{\left(\mu \omega^{2} / 2\right)\left(l^{2}-r^{2}\right) / \mathrm{A}}{\frac{\mathrm{d}(\delta)}{\mathrm{d} r}}$
$\therefore \frac{\mathrm{d}(\delta)}{\mathrm{d} r}=\frac{1}{Y A} \frac{\mu \omega^{2}}{2}\left(l^{2}-r^{2}\right)$
$\therefore \mathrm{d}(\delta)=\frac{1}{Y A} \frac{\mu \omega^{2}}{2}\left(l^{2}-r^{2}\right) d r$
$\therefore \delta=\frac{1}{Y A} \frac{\mu \omega^{2}}{2} \int_{0}^{l}\left(l^{2}-r^{2}\right) d r$
$=\frac{1}{Y A} \frac{\mu \omega^{2}}{2}\left[l^{3}-\frac{l^{3}}{3}\right]=\frac{1}{3 Y A} \mu \omega^{2} l^{3}=\frac{1}{3 Y A} \mu \omega^{2} l^{2}$
The total change in length is $2 \delta=\frac{2}{3 Y A} \mu \omega^{2} l^{2}$
9.27 An equilateral triangle $\mathrm{ABC}$ is formed by two $\mathrm{Cu}$ rods $\mathrm{AB}$ and $\mathrm{BC}$ and one $\mathrm{Al}$ rod. It is heated in such a way that temperature of each rod increases by $\Delta T$. Find change in the angle $A B C$. [Coeff. of
linear expansion for $\mathrm{Cu}$ is $\alpha_{1}$, Coeff. of linear expansion for $\mathrm{Al}$ is $\alpha_{2}$ ]
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Answer: Let $l_{1}=\mathrm{AB}, l_{2}=\mathrm{AC}, l_{3}=\mathrm{BC}$
$\cos \theta=\frac{l_{3}{ }^{2}+l_{1}{ }^{2}-l_{2}{ }^{2}}{2 l_{3} l_{1}}$
Or, $2 l_{3} l_{1} \cos \theta=l_{3}{ }^{2}+l_{1}{ }^{2}-l_{2}{ }^{2}$
Differenciating
$2\left(l_{3} d l_{1}+l_{1} d l_{3}\right) \cos \theta-2 l_{1} l_{3} \sin \theta d \theta$
$$ =2 l_{3} d l_{3}+2 l_{1} d l_{3}+2 l_{1} \alpha_{1}-2 l_{2} \alpha_{2} $$
Now, $\quad d l_{1}=l_{1} \alpha_{1} \Delta t$
$$ d l_{2}=l_{2} \alpha_{1} \Delta t $$
$$ d l_{3}=l_{3} \alpha_{2} \Delta t $$
and $l_{1}=l_{2}=l_{3}=l$
$\left(l^{2} \alpha_{1} \Delta t+l^{2} \alpha_{1} \Delta t\right) \cos \theta+l^{2} \sin \theta d \theta=l^{2} \alpha_{1} \Delta t+l^{2} \alpha_{1} \Delta t-l^{2} \alpha_{2} \Delta t$
$\sin \theta d \theta=2 \alpha_{1} \Delta t(1-\cos \theta)-\alpha_{2} \Delta t$
Putting $\theta=60^{\circ}$
$$ \begin{gathered} d \theta \frac{\sqrt{3}}{2}=2 \alpha_{1} \Delta t \times(1 / 2)-\alpha_{2} \Delta t \ =\left(\alpha_{1}-\alpha_{2}\right) \Delta t \end{gathered} \begin{aligned} & \text { Or, } d \theta=\frac{2\left(\alpha_{1}-\alpha_{2}\right) \Delta t}{\sqrt{3}} \end{aligned} $$
9.28 In nature, the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus the vertical through the centre of gravity does not fall within the base. The elastic torque caused because of this bending about the central axis of the tree is given by $\frac{Y \pi r^{4}}{4 R} . Y$ is the Young’s modulus, $r$ is the radius of the trunk and $R$ is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.
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Answer: When the tree is about to buckle
$$ W d=\frac{Y \pi r^{4}}{4 R} $$
If $R \quad h$, then the centre of gravity is at a height $l \quad \frac{1}{2} h$ from the ground.
From $\Delta \mathrm{ABC}$
$$ R^{2} \quad(R-d)^{2}+\left(\frac{1}{2} h\right)^{2} $$ Solution: N/A
If $d \quad R$
$R^{2} \quad R^{2}-2 R d+\frac{1}{4} h^{2}$
$\therefore d=\frac{h^{2}}{8 R}$
If $w_{0}$ is the weight/volume
$$ \begin{aligned} & \frac{Y \pi r^{4}}{4 R}=w_{0}\left(\pi r^{2} h\right) \frac{h^{2}}{8 R} \ & \Rightarrow h \quad\left(\frac{2 Y}{w_{o}}\right)^{1 / 3} r^{2 / 3} \end{aligned} $$
9.29 A stone of mass $m$ is tied to an elastic string of negligble mass and spring constant $k$. The unstretched length of the string is $L$ and has negligible mass. The other end of the string is fixed to a nail at a point $P$. Initially the stone is at the same level as the point $P$. The stone is dropped vertically from point $P$.
(a) Find the distance $y$ from the top when the mass comes to rest for an instant, for the first time.
(b) What is the maximum velocity attained by the stone in this drop?
(c) What shall be the nature of the motion after the stone has reached its lowest point?
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Answer: (a) Till the stone drops through a length $L$ it will be in free fall. After that the elasticity of the string will force it to a SHM. Let the stone come to rest instantaneously at $y$.
The loss in P.E. of the stone is the P.E. stored in the stretched string.
$$ \begin{aligned} & m g y=\frac{1}{2} k(y-L)^{2} \ & \text { Or, } m g y=\frac{1}{2} k y^{2}-k y L+\frac{1}{2} k L^{2} \ & \text { Or, } \frac{1}{2} k y^{2}-(k L+m g) y+\frac{1}{2} k L^{2}=0 \ & y=\frac{(k L+m g) \pm \sqrt{(k L+m g)^{2}-k^{2} L^{2}}}{k} \end{aligned} $$
$$ =\frac{(k L+m g) \pm \sqrt{2 m g k L+m^{2} g^{2}}}{k} $$
Retain the positive sign.
$\therefore y=\frac{(k L+m g)+\sqrt{2 m g k L+m^{2} g^{2}}}{k}$
(b) The maximum velocity is attained when the body passes, through the “equilibrium, position” i.e. when the instantaneous acceleration is zero. That is $m g-k x=0$ where $x$ is the extension from $\mathrm{L}$ :
$\Rightarrow m g=k x$
Let the velecity be $v$. Then
$\frac{1}{2} m v^{2}+\frac{1}{2} k x^{2}=m g(L+x)$
$\frac{1}{2} m v^{2}=m g(L+x)-\frac{1}{2} k x^{2}$
Now $m g=k x$
$$ \begin{aligned} x & =\frac{m g}{k} \ \therefore \frac{1}{2} m v^{2}= & m g\left(L+\frac{m g}{k}\right)-\frac{1}{2} k \frac{m^{2} g^{2}}{k^{2}} \ & =m g L+\frac{m^{2} g^{2}}{k}-\frac{1}{2} \frac{m^{2} g^{2}}{k} \end{aligned} $$
$$ \begin{aligned} & \frac{1}{2} m v^{2}=m g L+\frac{1}{2} \frac{m^{2} g^{2}}{k} \ & \therefore v^{2}=2 g L+m g^{2} / k \ & v=\left(2 g L+m g^{2} / k\right)^{1 / 2} \end{aligned} $$
(c) Consider the particle at an instantaneous position $y$. Then
$$ \begin{aligned} & \frac{m d^{2} y}{d t^{2}}=m g-k(y-L) \ & \Rightarrow \frac{d^{2} y}{d t^{2}}+\frac{k}{m}(y-L)-g=0 \end{aligned} $$
Make a transformation of variables: $z=\frac{k}{m}(y-L)-g$
Then $\frac{d^{2} z}{d t^{2}}+\frac{k}{m} z=0$
$\therefore z=A \cos (\omega t+\phi)$ where $\omega=\sqrt{\frac{k}{m}}$
$\Rightarrow y=\left(L+\frac{m}{k} g\right)+A^{\prime} \cos (\omega t+\phi)$
Thus the stone performs SHM with angular frequency $\omega$ about the point
$y_{0}=L+\frac{m}{k} g$
