14.6 Force law for simple harmonic motion

Using Newton’s second law of motion, and the expression for acceleration of a particle undergoing SHM (Eq. 14.11), the force acting on a particle of mass \(m\) in SHM is
\(
\begin{aligned}
F(t) & =m a \\
& =-m \omega^2 x(t)
\end{aligned}
\)
i.e., \(F(t)=-k x(t) \dots(14.13)\)
where \(k=m \omega^2 \dots(14.14a)\)
or \(\quad \omega=\sqrt{\frac{k}{m}} \dots(14.14b)\)
Where the constant \(k=m \omega^2\) is called the force constant or spring constant. The resultant force on the particle is zero when it is at the centre of oscillation. The centre of oscillation is, therefore, the equilibrium position. A force which takes the particle back towards the equilibrium position is called a restoring force in SHM.
Note that the force in Eq. (14.13) is linearly proportional to \(x(t)\). A particle oscillating under such a force is, therefore, calling a linear harmonic oscillator. In the real world, the force may contain small additional terms proportional to \(x^2, x^3\), etc. These then are called non-linear oscillators.

Example 1: The resultant force acting on a particle executing simple harmonic motion is \(4 \mathrm{~N}\) when it is \(5 \mathrm{~cm}\) away from the centre of oscillation. Find the spring constant.

Solution:

The simple harmonic motion is defined as
\(
F=-k x .
\)
The spring constant is \(k=\left|\frac{F}{x}\right|\)
\(
=\frac{4 \mathrm{~N}}{5 \mathrm{~cm}}=\frac{4 \mathrm{~N}}{5 \times 10^{-2} \mathrm{~m}}=80 \mathrm{~N} \mathrm{~m}^{-1} \text {. }
\)

Example 2: Two identical springs of spring constant \(k\) are attached to a block of mass \(m\) and to fixed supports, as shown in Fig. 14.14. Show that when the mass is displaced from its equilibrium position on either side, it executes a simple harmonic motion. Find the period of oscillations.

Solution:

Let the mass be displaced by a small distance \(x\) to the right side of the equilibrium position, as shown in Fig. 14.15. Under this situation, the spring on the left side gets elongated by a length equal to \(x\) and that on the right side gets compressed by the same length. 

The forces acting on the mass are then,
\(\begin{aligned} F_1=-k x & \text { (force exerted by the spring on } \\ & \text { the left side, trying to pull the } \\ & \text { mass towards the mean } \\ & \text { position) }\end{aligned}\)
\(F_2=-k x\) (force exerted by the spring on the right side, trying to push the mass towards the mean position)
The net force, \(F\), acting on the mass is then given by,
\(
F=-2 k x
\)
Hence the force acting on the mass is proportional to the displacement and is directed towards the mean position; therefore, the motion executed by the mass is simple harmonic. The time period of oscillations is,
\(
T=2 \pi \sqrt{\frac{m}{2 k}}
\)

Example 3: All the surfaces shown in the figure below are frictionless. The mass of the car is \(M\), that of the block is \(m\) and the spring has spring constant \(k\). Initially, the car and the block are at rest and the spring is stretched through a length \(x_0\) when the system is released. (a) Find the amplitudes of the simple harmonic motion of the block and of the car as seen from the road. (b) Find the time period(s) of the two simple harmonic motions.

Solution:

Let \(x_1\) and \(x_2\) be the amplitudes of oscillation of masses \(m\) and \(M\) respectively.
(a) As the centre of mass should not change during the motion, we can write:
\(
m x_1=M x_2 \dots(1)
\)
Let \(k\) be the spring constant. By conservation of energy, we have:
\(
\frac{1}{2} k x_0^2=\frac{1}{2} k\left(x_1+x_2\right)^2 \ldots(2)
\)
where \(\mathrm{x}_0\) is the length to which spring is stretched.
From equation (2) we have,
\(
x_0=x_1+x_2
\)
On substituting the value of \(x_2\) from equation (1) in equation (2), we get:
\(
\begin{aligned}
& x_0=x_1+\frac{m x_1}{M} \\
& \Rightarrow x_0=\left(1+\frac{m}{M}\right) x_1 \\
& \Rightarrow x_1=\left(\frac{M}{M+m}\right) x_0
\end{aligned}
\)
Now,\(x_2=x_0-x_1\)
On substituting the value of \(x_1\) from above equation, we get:
\(
\begin{aligned}
& \Rightarrow x_2=x_0\left[1-\frac{M}{M+m}\right] \\
& \Rightarrow x_2=\frac{m x_0}{M+m}
\end{aligned}
\)
Thus, the amplitude of the simple harmonic motion of a car, as seen from the road is \(\frac{m x_0}{M+m}\)

(b) At any position,
Let \(v_1\) and \(v_2\) be the velocities.
Using the law of conservation of energy we have,
\(
\frac{1}{2} M v^2+\frac{1}{2} m\left(v_1-v_2\right)^2+\frac{1}{2} k\left(x_1+x_2\right)^2=\text { constant } . . \text { (3) }
\)
Here, \(\left(v_1-v_2\right)\) is the absolute velocity of mass \(m\) as seen from the road.

Now, from the principle of conservation of momentum, we have:
\(
\begin{aligned}
& \mathrm{Mx}_2=\mathrm{mx}_1 \\
& \Rightarrow x_1=\left(\frac{M}{m}\right) x_2 \ldots \text { (4) } \\
& M v_2=m\left(v_1-v_2\right) \\
& \Rightarrow\left(v_1-v_2\right)=\left(\frac{M}{m}\right) v_2 \ldots(5)
\end{aligned}
\)
Putting the above values in equation (3), we get:
\(
\begin{aligned}
& \frac{1}{2} M v_2^2+\frac{1}{2} m \frac{M^2}{m^2} v_2^2+\frac{1}{2} k x_2^2\left(1+\frac{M}{m}\right)^2=\text { constant } \\
& \therefore M\left(1+\frac{M}{m}\right) v_2^2+k\left(1+\frac{M}{m}\right) x_2^2=\text { constant } \\
& \Rightarrow M v_2^2+k\left(1+\frac{M}{m}\right) x_2^2=\text { constant }
\end{aligned}
\)
Taking the derivative of both sides, we get:
\(
\begin{aligned}
& M \times 2 v_2 \frac{d v_2}{d t}+k\left(\frac{M+m}{m}\right) 2 x_2 \frac{d x_2}{d t}=0 \\
& \Rightarrow m a_2+k\left(\frac{M+m}{m}\right) x_2=0\left[\text { because }, v_2=\frac{d x_2}{d t}\right] \\
& \frac{a_2}{x_2}=\frac{-k(M+m)}{M m}=\omega^2 \\
& \therefore \omega=\sqrt{\frac{k(M+m)}{M m}}
\end{aligned}
\)

Example 4: Assume that a tunnel is dug across the earth (radius \(=R\) ) passing through its centre. Find the time a particle takes to cover the length of the tunnel if (a) it is projected into the tunnel with a speed of \(\sqrt{g R}\) (b) it is released from a height \(R\) above the tunnel (c) it is thrown vertically upward along the length of tunnel with a speed of \(\sqrt{g R}\).

Solution:

Given: The radius of the earth is \(R\).
Let \(M\) be the total mass of the earth and \(\rho\) be the density.
Let the mass of the part of the earth having radius \(x\) be \(M\).
\(
\begin{aligned}
& \therefore \frac{M^{\prime}}{M}=\frac{\rho \times \frac{4}{3} \pi x^3}{\rho \times \frac{4}{3} \pi R^3}=\frac{x^3}{R^3} \\
& \Rightarrow M^{\prime}=\frac{M x^3}{R^3}
\end{aligned}
\)
Force on the particle is calculated as,
\(
\begin{aligned}
& F_x=\frac{G M^{\prime} m}{x^2} \\
& =\frac{G M m}{R^3} x \ldots(1)
\end{aligned}
\)
Now, acceleration \(\left(a_x\right)\) of mass \(M^{\prime}\) at that position is given by,
\(
\begin{aligned}
& a_x=\frac{G M}{R^3} x \\
& \Rightarrow \frac{a_x}{x}=\omega^2=\frac{G M}{R^3}=\frac{g}{R}\left(\because g=\frac{G M}{R^2}\right)
\end{aligned}
\)
So, Time period of oscillation, \(T=2 \pi \sqrt{\left(\frac{R}{g}\right)}\)

(a) Velocity-displacement equation in S.H.M is written as,
\(
V=\omega \sqrt{\left(A^2-y^2\right)}
\)
where, \(A\) is the amplitude; and \(y\) is the displacement.
When the particle is at \(y=R\),
The velocity of the particle is \(\sqrt{g R}\) and \(\omega=\sqrt{\frac{g}{R}}\)
On substituting these values in the velocity-displacement equation, we get:
\(
\begin{aligned}
& \sqrt{g R}=\sqrt{\frac{g}{R}} \sqrt{A^2-R^2} \\
& \Rightarrow R^2=A^2-R^2 \\
& \Rightarrow A=\sqrt{2 R}
\end{aligned}
\)
Let \(t_1\) and \(t_2\) be the time taken by the particle to reach the positions \(\mathrm{X}\) and \(\mathrm{Y}\).
Now, the phase of the particle at point \(X\) will be greater than \(\frac{\pi}{2}\) but less than \(\pi\)
Also, the phase of the particle on reaching \(Y\) will be greater than \(\pi\) but less than \(\frac{3 \pi}{2}\)
Displacement-time relation is given by,
\(y=A \sin \omega t\)
Substituting \(y=R\) and \(\mathrm{A}=\sqrt{2 R}\), in the above relation, we get:
\(
\begin{aligned}
& R=\sqrt{2} R \sin \omega t_1 \\
& \Rightarrow \omega t_1=\frac{3 \pi}{4}
\end{aligned}
\)
Also,
\(
\begin{aligned}
& R=\sqrt{2} R \sin \omega t_2 \\
& \Rightarrow \omega t_2=\frac{5 \pi}{4} \\
& \text { So, } \omega\left(t_2-t_1\right)=\frac{\pi}{2} \\
& \Rightarrow t_2-t_1=\frac{\pi}{2 \omega}=\frac{\pi}{2\left(\sqrt{\frac{g}{R}}\right)}
\end{aligned}
\)

Time taken by the particle to travel from \(X\) to \(Y\) :
\(
t_2-t_1=\frac{\pi}{2 \omega}=\frac{\pi}{2} \sqrt{\frac{R}{g}} \mathrm{~s}
\)
(b) When the body is dropped from a height \(R\)
Using the principle of conservation of energy, we get:
Change in P.E. = Gain in K.E.
\(
\begin{aligned}
& \Rightarrow \frac{G M m}{R}-\frac{G M m}{2 R}=\frac{1}{2} m v^2 \\
& \Rightarrow v=\sqrt{(g R)}
\end{aligned}
\)
As the velocity is the same as that at \(X\), the body will take the same time to travel \(X Y\).
(c) The body is projected vertically upwards from the point \(\mathrm{X}\) with a velocity \(\sqrt{g R}\). Its velocity becomes zero as it reaches the highest point.
The velocity of the body as it reaches \(X\) again will be,
\(
v=\sqrt{(g R)}
\)
Hence, the body will take same time i.e.
\(
\frac{\pi}{2} \sqrt{\left(\frac{R}{g}\right)} \text { s to travel } X Y
\)

Example 5: Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance \(R / 2\) from the earth’s centre where \(R\) is the radius of the earth. The wall of the tunnel is frictionless. (a) Find the gravitational force exerted by the earth on a particle of mass \(m\) placed in the tunnel at a distance \(x\) from the centre of the tunnel. (b) Find the component of this force along the tunnel and perpendicular to the tunnel. (c) Find the normal force exerted by the wall on the particle. (d) Find the resultant force on the particle. (e) Show that the motion of the particle in the tunnel is simple harmonic and find the time period.

Solution:

If \(\rho\) is the density of the earth, then the mass of the earth \((M)\) is given by,
\(
M=\frac{4}{3} \pi R^3 \rho
\)
Similarly, mass \(\left(M^{\prime}\right)\) of the part of earth having radius \((x)\) is given by, \(M^{\prime}=\frac{4}{3} \pi x_1{ }^3 \rho\)
\(
M^{\prime}=\left(\frac{M}{R^3}\right) x_1^3
\)
(a) Let \(F\) be the gravitational force exerted by the earth on the particle of mass \(m\). Then, its value is given by,
\(
F=\frac{G M^{\prime} m}{x_1^2}
\)
Substituting the value of \(\mathrm{M}^{\prime}\) in the above equation, we get:
\(
\begin{aligned}
& F=\frac{G M m}{R^3} \frac{x_1^3}{x_1^2} \\
& =\frac{G M m}{R^3} x_1=\frac{G M m}{R^3} \sqrt{x^2+\left(\frac{R^2}{4}\right)}
\end{aligned}
\)

(b)

\(
\begin{aligned}
& F_y=F \cos \theta \\
& =\frac{G M m x_1}{R^3} \frac{x}{x_1}=\frac{G M m x}{R^3} \\
& F_x=F \sin \theta \\
& =\frac{G M m x_1}{R^3} \frac{R}{2 x_1}=\frac{G M m}{2 R^2}
\end{aligned}
\)

(c)
\(
F_x=\frac{G M m}{R^2}
\)
\(\because\) Normal force exerted by the wall \(N=F_X\)
(d) The resultant force is \(\frac{G M m x}{R^3}\)
(e) Acceleration = Driving force/mass
\(
\begin{aligned}
& =\frac{G M m x}{R^3 m} \\
& =\frac{G M x}{R^3} \\
& \Rightarrow a \propto x \quad \text { (the body executes S.H.M.) } \\
& \frac{a}{x}=\omega^2=\frac{G M}{R^3} \\
& \Rightarrow \omega=\sqrt{\frac{G m}{R^3}} \\
& \Rightarrow T=2 \pi \sqrt{\frac{R^3}{G M}}
\end{aligned}
\)

Example 6: A simple pendulum of length \(l\) is suspended through the ceiling of an elevator. Find the time period of small oscillations if the elevator (a) is going up with an acceleration \(a_0\) (b) is going down with an acceleration \(a_0\) and (c) is moving with a uniform velocity.

Solution:

(a)
The length of the simple pendulum is \(l\).
Let \(x\) be the displacement of the simple pendulum.
From the diagram, the driving forces \(f\) is given by,
\(
f=m\left(g+a_0\right) \sin \theta \dots(1)
\)
Acceleration (a) of the elevator is given by,
\(
\begin{aligned}
& a=\frac{f}{m} \\
& =\left(g+a_0\right) \sin \theta \\
& =\left(g+a_0\right) \frac{x}{l}\left(\text { From the diagram } \sin \theta=\frac{x}{l}\right)
\end{aligned}
\)
[ when \(\theta\) is very small, \(\sin \theta \rightarrow \theta=x / l\) ]
\(
\therefore a=\left(\frac{g+a_0}{l}\right) x \dots(2)
\)
As the acceleration is directly proportional to displacement, the pendulum executes S.H.M. Comparing equation (2) with the expression \(a=\omega^2 x\), we get:
\(
\omega^2=\frac{g+a_0}{l}
\)
Thus, the time period of small oscillations when the elevator is going upward \((T)\) will be:
\(
T=2 \pi \sqrt{\frac{l}{g+a_0}}
\)
(b)

When the elevator moves downwards with acceleration \(a_0\),
Driving force \((F)\) is given by,
\(
F=m\left(g-a_0\right) \sin \theta
\)
On comparing the above equation with the expression, \(F=m a_1\) Acceleration, \(a=\left(g-a_0\right) \sin \theta=\frac{\left(g-a_0\right) x}{l}=\omega^2 x\) Time period of the elevator when it is moving downward \(\left(T^{\prime}\right)\) is given by,
\(
T^{\prime}=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{l}{g-a_0}}
\)
(c) When the elevator moves with uniform velocity, i.e. \(a_0=0\),
For a simple pendulum, the driving force \((F)\) is given by,
\(
F=\frac{m g x}{l}
\)
Comparing the above equation with the expression, \(\mathrm{F}=\mathrm{ma}\), we get: \(a=\frac{g x}{l} \Rightarrow \frac{x}{a}=\frac{l}{g} T=2 \pi \sqrt{\frac{\text { displacement }}{\text { Acceleration }}}\)
\(
=2 \pi \sqrt{\frac{l}{g}}
\)

Example 7: A simple pendulum of length \(l\) is suspended from the ceiling of a car moving with a speed \(v\) on a circular horizontal road of radius \(r\). (a) Find the tension in the string when it is at rest with respect to the car. (b) Find the time period of small oscillation.

Solution:

It is given that a car is moving with speed \(v\) on a circular horizontal road of radius \(r\).
(a) Let \(T\) be the tension in the string.

\(
\text { According to the free body diagram, the value of } T \text { is given as, }
\)
\(
\begin{aligned}
& T=\sqrt{(m g)^2+\left(\frac{m v^2}{r}\right)^2} \\
& =m \sqrt{g^2+\frac{v^4}{r^2}}=m a
\end{aligned}
\)
where acceleration, \(a=\sqrt{g^2+\frac{v^4}{r^2}}\)
The time period \((T)\) is given by ,
\(
T=2 \pi \sqrt{\frac{l}{g}}
\)
On substituting the respective values, we have:
\(
T=2 \pi \sqrt{\frac{l}{\left(g^2+\frac{v^4}{r^2}\right)^{1 / 2}}}
\)

Example 8: Find the time period of small oscillations of the following systems. (a) A metre stick suspended through the \(20 \mathrm{~cm}\) mark. (b) A ring of mass \(m\) and radius \(r\) suspended through a point on its periphery. (c) A uniform square plate of the edge \(a\) suspended through a corner. (d) A uniform disc of mass \(m\) and radius \(r\) suspended through a point \(r / 2\) away from the centre.

Solution:

\(
\text { (a) Moment Of inertia }(I) \text { about the point } X \text { is given by, }
\)
\(
\begin{aligned}
& \mathrm{I}=\mathrm{I}_{\mathrm{C} . G}+m h^2 \\
& =\frac{m l^2}{12}+m h^2 \\
& =\frac{m l^2}{12}+m(0.3)^2 \\
& =m\left(\frac{1}{12}+0.09\right) \\
& =m\left(\frac{1+1.08}{12}\right) \\
& =m\left(\frac{2.08}{12}\right)
\end{aligned}
\)
The time period \((T)\) is given by,
\(T=2 \pi \sqrt{\frac{I}{m g l}}\) where \(I=\) the moment of inertia, and
\(l=\) distance between the centre of gravity and the point of suspension.
On substituting the respective values in the above formula, we get: \(T=2 \pi \sqrt{\frac{2.08 m}{m \times 12 \times 9.8 \times 0.3}}\)
\(=1.52 \mathrm{~s}\)
(b)
\(
\text { Moment Of inertia }(I) \text { about A is given as, }
\)
\(
I=I_{C . G .}+m r^2=m r^2+m r^2=2 m r^2
\)
The time period \((T)\) will be,
\(
T=2 \pi \sqrt{\frac{I}{m g l}}
\)
On substituting the respective values in the above equation, we have:
\(
\begin{aligned}
& T=2 \pi \sqrt{\frac{2 m r^2}{m g r}} \\
& =2 \pi \sqrt{\frac{2 r}{g}}
\end{aligned}
\)
(c) Let \(I\) be the moment of inertia of a uniform square plate suspended through a corner.
\(
I=m\left(\frac{a^2+a^2}{3}\right)=\frac{2 m}{3} a^2
\)
In the
\(
\begin{aligned}
& \triangle \mathrm{ABC}, l^2+l^2=a^2 \\
& \therefore l=\frac{a}{\sqrt{2}} \\
& \Rightarrow T=2 \pi \sqrt{\frac{I}{m g l}} \\
& =2 \pi \sqrt{\frac{2 m a^2}{3 m g l}} \\
& =2 \pi \sqrt{\frac{2 a^2}{3 g a \sqrt{2}}} \\
& =2 \pi \sqrt{\frac{\sqrt{8} a}{3 g}}
\end{aligned}
\)
(d)
We know
\(
h=\frac{r}{2}
\)
Distance between the \(C . G\). and suspension point, \(l=\frac{r}{2}\)
Moment of inertia about \(A\) will be:
\(
\begin{aligned}
& \mathrm{I}=\mathrm{I}_{\text {C.G. }}+m h^2 \\
= & \frac{m r^2}{2}+m\left(\frac{r}{2}\right)^2 \\
= & m r^2\left(\frac{1}{2}+\frac{1}{4}\right)=\frac{3}{4} m r^2
\end{aligned}
\)
Time period \((T)\) will be,
\(
\begin{aligned}
& T=2 \pi \sqrt{\frac{I}{m g l}} \\
& =2 \pi \sqrt{\frac{3 m r^2}{4 m g l}}=2 \pi \sqrt{\frac{3 r}{2 g}}
\end{aligned}
\)

Example 9: A closed circular wire hung on a nail in a wall undergoes small oscillations of amplitude \(2^{\circ}\) and time period \(2 \mathrm{~s}\). Find (a) the radius of the circular wire, (b) the speed of the particle farthest away from the point of suspension as it goes through its mean position, (c) the acceleration of this particle as it goes through its mean position and (d) the acceleration of this particle when it is at an extreme position. Take \(g=\pi^2 \mathrm{~m} \mathrm{~s}^{-2}\).

Solution:

It is given that:
Time period of oscillation, \(T=2 \mathrm{~s}\)
Acceleration due to gravity, \(g=\pi^2 \mathrm{~ms}^{-2}\)
Let \(I\) be the moment of inertia of the circular wire having mass \(m\) and radius \(r\).
(a) Time period of the compound pendulum (T) is given by,
\(
\begin{aligned}
& T=2 \sqrt{\frac{I}{m g l}}=2 \sqrt{\frac{I}{m g r}} \\
& (\because l=r) \ldots(1)
\end{aligned}
\)
The moment of inertia about the point of suspension is calculated as,
\(
\mathrm{I}=\mathrm{mr}^2+\mathrm{mr}^2=2 \mathrm{mr}^2
\)
On substituting the value of the moment of inertia \(I\) in equation (1), we get:
\(
\begin{aligned}
& T=2 \pi \sqrt{\frac{2 m r^2}{m g r}}=2 \pi \sqrt{\frac{2 r}{g}} \\
& \Rightarrow \frac{2}{2 \pi}=\sqrt{\frac{2 r}{g}} \\
& \Rightarrow \frac{2 r}{g}=\frac{1}{\pi^2} \\
& \Rightarrow r=\frac{g}{2 \pi^2} \\
& =0.5 m=50 \mathrm{~cm}
\end{aligned}
\)
(b) From the energy equation, we have:
\(
\begin{aligned}
& \frac{1}{2} I \omega^2-0=\operatorname{mgr}(1-\cos \theta) \\
& \frac{1}{2} I \omega^2-0=\operatorname{mgr}\left(1-\cos 2^{\circ}\right) \\
& \Rightarrow\left(\frac{1}{2}\right) 2 m r^2 \cdot \omega^2=m g r\left(1-\cos 2^{\circ}\right)\left(\because I=2 m r^2\right) \\
& \Rightarrow \omega^2=\frac{g}{r}\left(1-\cos 2^{\circ}\right)
\end{aligned}
\)
On substituting the value of \(g\) and \(r\) in the above equation, we get: \(\omega=0.11 \mathrm{rad} / \mathrm{s}\)
\(
\Rightarrow v=\omega \times 2 r=11 \mathrm{cms}^{-1}
\)
(c) The acceleration is found to be centripetal at the extreme position. Centripetal acceleration at the extreme position \(\left(a_n\right)\) is given by,
\(
a_{\mathrm{n}}=\omega^2(2 r)=(0.11) \times 100=12 \mathrm{~cm} / \mathrm{s}^2
\)
The direction of \(a_n\) is towards the point of suspension.
(d) The particle has zero centripetal acceleration at the extreme position. However, the particle will still have acceleration due to the S.H.M.
Angular frequency \((\omega)\) is given by ,
\(
\begin{aligned}
& \omega=\frac{2 \pi}{T} \\
& =\frac{2 \pi}{2}=3.14
\end{aligned}
\)
\(\therefore\) Angular Acceleration \((a)\) at the extreme position is given as,
\(
\begin{aligned}
& \alpha=\omega^2 \theta \\
& \alpha=\omega^2 2^{\circ}=\pi^2 \times \frac{2 \pi}{180} \\
& =\frac{2 \pi^3}{180}\left[1^{\circ}=\frac{\pi}{180} \text { radian }\right]
\end{aligned}
\)
Thus, tangential acceleration
\(
\begin{aligned}
& =\alpha(2 r)=\left(\frac{2 \pi^3}{180}\right) \times 100 \\
& =34 \mathrm{~cm} / \mathrm{s}^2
\end{aligned}
\)