Simple Harmonic Motion HC Verma PDF

# CHAPTER 12 ## SIMPLE HARMONIC MOTION ### 12.1 SIMPLE HARMONIC MOTION When a body repeats its motion after regular time intervals we say that it is in harmonic motion or periodic motion. The time interval after which the motion is repeated is called the time period. If a body moves to and fro o...

# CHAPTER 12 ## SIMPLE HARMONIC MOTION ### 12.1 SIMPLE HARMONIC MOTION When a body repeats its motion after regular time intervals we say that it is in harmonic motion or periodic motion. The time interval after which the motion is repeated is called the time period. If a body moves to and fro on the same path, it is said to perform oscillations. Simple harmonic motion (SHM) is a special type of oscillation in which the particle oscillates on a straight line, the acceleration of the particle is always directed towards a fixed point on the line and its magnitude is proportional to the displacement of the particle from this point. This fixed point is called the centre of oscillation. Taking fixed point as the origin and the line of motion as the X-axis, we can write the defining equation of a simple harmonic motion as $a = -ω²x$ (12.1) where $w²$ is a positive constant. If x is positive, a is negative and if x is negative, a is positive. This means that the acceleration is always directed towards the centre of oscillation. If we are looking at the motion from an inertial frame, $a=F/m$. The defining equation (12.1) may thus be written $\frac{F}{m} = -w²x$ or, $F = -mw²x$ or, $F = -kx$. (12.2) We can use equation (12.2) as the definition of SHM. A particle moving on a straight line executes simple harmonic motion if the resultant force acting on it is directed towards a fixed point on the line and is proportional to the displacement of the particle from this fixed point. The constant $k = mw²$ 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. Equation (12.2) represents a restoring force which is linear. Figure (12.1) shows the linear restoring force graphically. **Example 12.1** The resultant force acting on a particle executing simple harmonic motion is 4 N when it is 5 cm away from the centre of oscillation. Find the spring constant. **Solution:** The simple harmonic motion is defined as $F=-kx$. The spring constant is $k = \frac{F}{x}$ $ = \frac{4 N}{5 cm}$ $ = \frac{4 N}{5 × 10^{-2} m}$ $ = 80 N m^{-1}$. ### 12.2 QUALITATIVE NATURE OF ### SIMPLE HARMONIC MOTION Let us consider a small block of mass m placed on a smooth horizontal surface and attached to a fixed wall through a spring as shown in figure (12.2). Let the spring constant of the spring be k. [Diagram image description: A block is attached to a spring, the spring is attached to a wall. The block is pulled to the right, extending the spring, and labeled P. At rest, before the block is pulled, the block is labeled O. The block is pulled to a point labeled Q to the left, compressing the spring. The center point of the spring is labeled O.] ### 12.3 EQUATION OF MOTION OF ### A SIMPLE HARMONIC MOTION Consider a particle of mass m moving along the X-axis. Suppose, a force F = kx acts on the particle where k is a positive constant and x is the displacement of the particle from the assumed origin. The particle then executes a simple harmonic motion with the centre of oscillation at the origin. We shall calculate the displacement x and the velocity v as a function of time. [Diagram image description: A horizontal x axis. A particle is located at x0. The particle moves to the right along the x axis to x. The particle then moves left along the x axis to x0 as the particle's velocity increases to v0 in the positive direction.] Suppose the position of the particle at t = 0 is x0 and its velocity is v0. Thus, at t = 0, x = x0, and v = v0. The acceleration of the particle at any instant is $a = \frac{F}{m}$ $ = \frac{k}{m} x = -w²x$, where $w = \sqrt{\frac{k}{m}}$. or, $\frac{dv}{dt} = -w²x$ or, $dv = -w²x dx$ or, $v dv = -w²x dx$. The velocity of the particle is v0 when the particle is at x = x0. It becomes v when the displacement becomes x. We can integrate the above equation and write $\int_{v0}^v v dv = \int_{x_0}^x -w²x dx$ or, $\left[ \frac{v²}{2}\right]_{v0}^v = \left[- w² \frac{x²}{2}\right]_{x_0}^x$ or, $v² = v_0² + w²(x_0² - x²)$ or, $v² = (v_0² + w²x_0²)- w²x²$ or, $v = \sqrt{(v_0² + w²x_0²)- w²x²}$ or, $v = w\sqrt{(\frac{v_0²}{w²}+x_0²)- x^2}$ Writing $\frac{v_0²}{w²} + x_0² = A²$ the above equation becomes $v = w \sqrt{A² - x²}$. (12.4) We can write this equation as $\frac{dx}{dt} = w \sqrt{A² - x²}$ or, $\frac{dx}{\sqrt{A² - x²}} = w dt$. At time t = 0 the displacement is x = x0 and at time t the displacement becomes x. The above equation can be integrated as $\int_{x0}^x \frac{dx}{\sqrt{A² - x²}} = \int_{0}^t w dt$ or, $\left[ sin^{-1}\frac{x}{A} \right]_{x0}^x = [wt]_0^t$ or, $sin^{-1}\frac{x}{A} - sin^{-1}\frac{x_0}{A} = wt$ Writing $sin^{-1}\frac{x_0}{A} = δ$, this becomes $sin^{-1}\frac{x}{A} = wt+δ$ or, $x = A sin(wt+δ)$. (12.6) The velocity at time t is $v=\frac{dx}{dt}= A w cos(wt+δ)$. (12.7) ### TERMS ASSOCIATED WITH SIMPLE ### HARMONIC MOTION #### Amplitude Equation (12.6) gives the displacement of a particle in simple harmonic motion. As sin(wt+δ) can take values between -1 and + 1, the displacement x can take values between -A and + A. This gives the physical significance of the constant A. It is the maximum displacement of the particle from the centre of oscillation, i.e. the amplitude of oscillation. #### Time Period A particle in simple harmonic motion repeats its motion after a regular time interval. Suppose the particle is at a position x and its velocity is v at a certain time t. After some time the position of the particle will again be x and its velocity will again be v in the same direction. This part of the motion is called one complete oscillation and the time taken in one complete oscillation is called the time period T. Thus, in figure (12.4) Q to P and then back to Q is a complete oscillation, R to P to Q to R is a complete oscillation, O to P to Q to O is a complete oscillation, etc. Both the position and the velocity (magnitude as well as direction) repeat after each complete oscillation. [Diagram image description: A horizontal x axis with a sine wave, the sine wave is labeled with points Q, R, O, and P.] If T be the time period, x should have same value at t and t+T. Thus, $sin(wt+δ)=sin[w(t+T)+δ]$. Now the velocity is (equation 12.7) v = A w cos(wt+δ). As the velocity also repeats its value after a time period, cos(wt+δ) = cos[w(t+T) + δ]. Both sin(wt+δ) and cos(wt+δ) will repeat their values if the angle (wt+δ) increases by 2π or its multiple. As T is the smallest time for repetition, w(t+T) + δ = (wt+δ) + 2π or, wT = 2π or, $T = \frac{2\pi}{w}$ Remembering that w = √km, we can write for the time period, $T = 2\pi\sqrt{\frac{m}{k}}$ (12.8) where k is the force constant and m is the mass of the particle. **Example 12.3** A particle of mass 200 g executes a simple harmonic motion. The restoring force is provided by a spring of spring constant 80 Nm¯¹. Find the time period. **Solution:** The time period is $T = 2\pi\sqrt{\frac{m}{k}}$. #### Frequency and Angular Frequency The reciprocal of time period is called the frequency. Physically, the frequency represents the number of oscillations per unit time. It is measured in cycles per second also known as herts and written in symbols as Hz. Equation (12.8) shows that the frequency is $v = \frac{1}{T}$ ... (12.9) $v = \frac{w}{2 \pi} = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}$ ... (12.10) The constant w is called the angular frequency. #### Phase The quantity $wt+8$ is called the phase. It determines the status of the particle in simple harmonic motion. If the phase is zero at a certain instant, x=A sin(wt+8) = 0 and v=Aw cos(wt+8) =A. This means that the particle is crossing the mean position and is going towards the positive direction. If the phase is π/2, we get x=A, v=0 so that the particle is at the positive extreme position. Figure (12.5) shows the status of the particle at different phases. [Diagram image description: A sine wave is drawn. Along the x axis the position of the particle is described at a time t = 0, wt = π/2, wt = π, wt = 3/2 π, and wt = 2π. At wt = 0 the particle crosses the x axis traveling in the positive direction. At wt = π/2 the particle is at it's maximum position on the positive x axis. wt = π the particle crosses the x axis moving in the negative direction. At wt = 3/2 π the particle is at it's maximum position on the negative x axis. At wt = 2π the particle crosses the x axis traveling in the positive direction.] We see that as time increases the phase increases. An increase of 2π brings the particle to the same status in the motion. Thus, a phase wt+8 is equivalent to a phase wt+8+2π. Similarly, a phase change of 4π, 6π, 8π, ... , etc., are equivalent to no phase change. Figure (12.6) shows graphically the variation of position and velocity as a function of the phase. [Diagram image description: A sinusoidal wave is drawn. The x axis is labeled phase. The y axis is labeled x. A second sine wave is drawn below the first one. The x axis is labeled phase The y axis is labeled v.] #### Phase Constant The constant δ appearing in equation (12.6) is called the phase constant. This constant depends on the choice of the instant t = 0. To describe the motion quantitatively, a particular instant should be called t=0 and measurement of time should be made in comparison with this instant and is going according to the convenience of the problem. Suppose we choose this instant. This instant may be chosen according to the convenience of the problem. Suppose we choose t=0 at an instant when the particle is passing through its mean position and is going towards the positive direction. The phase wt+δ should then be zero. As t=0 this means δ will be zero. The equation for displacement can then be written as x=Asinot. If we choose t=0 at an instant when the particle is at its positive extreme position, the phase is π/2 at this instant, Thus at + δ = π/2 and hence δ =π/2. The equation for the displacement is x = A sin(wt + π/2) or, x=Acosot. Any instant can be chosen as t = 0 and hence the phase constant can be chosen arbitrarily. Quite often we shall choose δ = 0 and write the equation for displacement as x = A sinot. Sometimes we may have to consider two or more simple harmonic motions together. The phase constant of any one can be chosen as δ = 0. The phase constants of the rest will be determined by the actual situation. The general equation for displacement may be written as x = A sin(wt+δ) $=A sin(wt+\frac{\pi}{2}+\delta')$ $=Acos(wt+\delta')$ where δ' is another arbitrary constant. The sine form and the cosine form are, therefore, equivalent. The value of phase constant, however, depends on the form chosen. **Example 12.4** A particle executes simple harmonic motion of amplitude A along the X-axis. At t = 0, the position of the particle is x = A/2 and it moves along the positive x-direction. Find the phase constant δ if the equation is written as x = A sin(wt + δ). **Solution:** We have x = A sin(wt + δ). At t=0, x=A/2. Thus, A/2 = A sind or, sind=1/2 or, $δ=π/6$ or $5π/6$. The velocity is $v=\frac{dx}{dt}= A w cos(wt + δ)$. At $t = 0$, $v = A w cοςδ$. Now, $cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$ and $cos\frac{5 \pi}{6} = -\frac{\sqrt{3}}{2}$ As v is positive at t = 0, δ must be equal to π/6. ### 12.5 SIMPLE HARMONIC MOTION AS A ### PROJECTION OF CIRCULAR MOTION Consider a particle P moving on a circle of radius A with a constant angular speed w (figure 12.7). Let us take the centre of the circle as the origin and two perpendicular diameters as the X and Y-axes. Suppose the particle P is on the X-axis at t = 0. The radius OP will make an angle wt with the X-axis at time t. Drop perpendicular PQ on X-axis and PR on Y-axis. The x and y-coordinates of the particle at time t are x = OQ = OP cos wt or, x = A cos wt and y = OR = OP sin wt (12.11) or, y = A sin wt. (12.12) [Diagram image description: A circle with a radius A is drawn. Point O is at the center of the circle. Point P is at the intersection of the circle and the radius. Point Q is the point of intersection of the x axis and the perpendicular line from point P to the x axis. Point R is the point of intersection of the y axis and the perpendicular line from point P to the y axis.] Equation (12.11) shows that the foot of perpendicular Q executes a simple harmonic motion on the X-axis. The amplitude is A and the angular frequency is w. Similarly, equation (12.12) shows that the foot of perpendicular R executes a simple harmonic motion on the Y-axis. The amplitude is A and the angular frequency is w. The phases of the two simple harmonic motions differ by π/2 [remember coswt = sin(wt+ π/2)]. Thus, the projection of a uniform circular motion on a diameter of the circle is a simple harmonic motion. ### 12.6 ENERGY CONSERVATION IN SIMPLE ### HARMONIC MOTION Simple harmonic motion is defined by the equation F = - kx. The work done by the force F during a displacement from x to x + dx is dW = F dx $= -kx dx$. The work done in a displacement from x = 0 to x is $W = \int_{0}^x (-kx)dx = -\frac{1}{2}kx^2$. Let U(x) be the potential energy of the system when the displacement is x. As the change in potential energy corresponding to a force is negative of the work done by this force, U(x) - U(0) = W = - $\frac{1}{2}kx^2$. Let us choose the potential energy to be zero when the particle is at the centre of oscillation x = 0. Then U(0) = 0 and U(x) = $\frac{1}{2}x^2$. This expression for potential energy is same as that for a spring and has been used so far in this chapter. As $k = mw^2$ , we can write $U(x) = \frac{1}{2}mw^2x^2$. (12.13) The displacement and the velocity of a particle executing a simple harmonic motion are given by x = A sin(wt+δ) and v = A w cos(wt+δ). The potential energy at time t is, therefore, $U = \frac{1}{2}mw^2x^2$ $= \frac{1}{2}mw^2A^2sin²(wt+δ)$. and the kinetic energy at time t is $K=\frac{1}{2}mv²$ $= \frac{1}{2}mA²w²cos²(wt+δ).$ The total mechanical energy at time t is E = U + K $= \frac{1}{2}mw²A²[sin²(wt+δ)+cos²(wt+δ)]$ $= \frac{1}{2}mw²A²$. (12.14) We see that the total mechanical energy at time t is independent of t. Thus, the mechanical energy remains constant as expected. As an example, consider a small block of mass m placed on a smooth horizontal surface and attached to a fixed wall through a spring of spring constant k (figure 12.8). [Diagram image description: A block is attached to a spring, the spring is attached to a wall. The block is pulled to the right, extending the spring, and labeled P. At rest, before the block is pulled, the block is labeled O. The block is pulled to a point labeled Q to the left, compressing the spring. The center point of the spring is labeled O.] ### 12.7 ANGULAR SIMPLE HARMONIC MOTION A body free to rotate about a given axis can make angular oscillations. For example, a hanging umbrella makes angular oscillations when it is slightly pushed aside and released. The angular oscillations are called angular simple harmonic motion if (a) there is a position of the body where the resultant torque on the body is zero, this position is the mean position θ = 0, (b) when the body is displaced through an angle θ from the mean position, a resultant torque acts which is proportional to the angle displaced, and (c) this torque has a sense (clockwise or anticlockwise) so as to bring the body towards the mean position. If the angular displacement of the body at an instant is θ, the resultant torque acting on the body in angular simple harmonic motion should be Γ = -kθ If the moment of inertia is I, the angular acceleration is $α = \frac{Γ}{I} = -\frac{k}{I}θ$ or, $\frac{d^2θ}{dt^2} = -w² θ$ (12.15) where w² = k/I. Equation (12.15) is identical to equation (12.3) except for the symbols. The linear displacement x in (12.3) is replaced here by the angular displacement θ. Thus, equation (12.15) may be integrated in the similar manner and we shall get an equation similar to (12.6), i.e. θ = θ₀ sin(wt + δ) (12.16) where θ₀ is the maximum angular displacement on either side. The angular velocity at time t is given by, $\frac{dθ}{dt} = θω cos(wt + δ)$. The time period of oscillation is $T = \frac{2\pi}{w}$ $= 2\pi\sqrt{\frac{I}{k}}$ (12.17) and the frequency of oscillation is $v = \frac{1}{T}$ $ = \frac{w}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{k}{I}}$ (12.18) (12.19) The quantity w = √k/I is the angular frequency. **Example 12.6** A body makes angular simple harmonic motion of amplitude π/10 rad and time period 0.05 s. If the body is at a displacement θ = π/10 rad at t = 0, write the equation giving the angular displacement as a function of time. **Solution:** Let the required equation be θ=θ₀sin(wt + δ). Here θ₀ = amplitude = π/10 rad $T = \frac{2\pi}{w}$ $= 0.05 s$ $w = \frac{2\pi}{0.05s} = 40 \pi s^{-1}$ so that $θ= \frac{\pi}{10} rad sin [(40π s¯*)t+δ]$. At $t=0$, $θ=\frac{\pi}{10} rad$. Putting in (i), $sin(δ) = \frac{\pi/10}{\pi/10}$ or, $sind =1$ or, $δ = \frac{\pi}{2}$. Thus by (i), $θ= \frac{\pi}{10} rad sin (40π s¯¹t)+ \frac{\pi}{2}$ $= \frac{\pi}{10} rad cos[(40π s¯¹t] $. ### 12.8 SIMPLE PENDULUM A simple pendulum consists of a heavy particle suspended from a fixed support through a light inextensible string. Simple pendulum is an idealised model. In practice, one takes a small metallic sphere and suspends it through a string. Figure (12.9) shows a simple pendulum in which a particle of mass m is suspended from the fixed support O through a light string of length l. The system can stay in equilibrium if the string is vertical. This is the mean or equilibrium position. If the particle is pulled aside and released, it oscillates in a circular arc with the centre at the point of suspension O. [Diagram image description: A right triangle is drawn. The length of the hypotenuse is labeled l. The angle of the triangle at the bottom is labeled 0. The hypotenuse is labeled with a point labeled "P" at the end. The side opposite the angle labeled 0 is labeled OQ and the side adjacent to the angle labeled 0 is labeled OP. The angle at point O is labeled with a point labeled O. The particle is labeled with a point labeled "P".] The position of the particle at any time can be described by the angle θ between the string and the vertical. The mean position or the equilibrium position corresponds to θ = 0. The particle makes pure rotation about the horizontal line OA (figure 12.9) which is perpendicular to the plane of motion. Let us see whether the motion of the particle is simple harmonic or not and find out its time period of oscillation. Let the particle be at P at a time t when the string makes an angle θ with the vertical (figure 12.10). Let OQ be the horizontal line in the plane of motion. Let PQ be the perpendicular to OQ. [Diagram image description A right triangle is drawn. The length of the hypotenuse is labeled l. The angle of the triangle at the bottom is labeled 0. The hypotenuse is labeled with a point labeled "P" at the end. The side opposite the angle labeled 0 is labeled OQ and the side adjacent to the angle labeled 0 is labeled OP. The angle at point O is labeled with a point labeled O. The particle is labeled with a point labeled "P". A vertical line is drawn from point P to point Q which is on the x-axis. The force of gravity is labeled with a vector labeled mg that acts on point P and points downward.] Forces acting on the particle are (a) the weight mg and (b) the tension T. The torque of T about OA is zero as it intersects OA. The magnitude of the torque of mg about OA is |Γ| = (mg) (OQ) = mg (OP) sine = mgl sine. Also, the torque tries to bring the particle back towards θ = 0. Thus, we can write Γ = - mgl sine. (12.21) We see that the resultant torque is not proportional to the angular displacement and hence the motion is not angular simple harmonic. However, if the angular displacement is small, sine is approximately equal to θ (expressed in radians) and equation (12.21) may be written as Γ = - mgl θ. (12.22) Thus, if the amplitude of oscillation is small, the motion of the particle is approximately angular simple harmonic. The moment of inertia of the particle about the axis of rotation OA is I = m(OP)² = ml². The angular acceleration is α = $\frac{Γ}{I}$ = $\frac{mglθ}{ml^2}$ = $-\frac{g}{l}θ$ or, $α = -w²θ$ where w² = g/l This is the equation of an angular simple harmonic motion. The constant w = √gl¯¹ represents the angular frequency. The time period is $T = \frac{2\pi}{w}$ $= 2\pi\sqrt{\frac{l}{g}}$. ... (12.23) **Example 12.7** Calculate the time period of a simple pendulum of length one meter. The acceleration due to gravity at the place is π²ms². **Solution:** The time period is $T = 2\pi\sqrt{\frac{l}{g}}$ $ = 2\pi \sqrt{\frac{ 1 m}{π^2 m s^{-2}}}$ = 2 s. #### Simple Pendulum as a Linear #### Simple Harmonic Oscillator If the amplitude of oscillation is small, the path of the particle is approximately a straight line and the motion can be described as a simple harmonic motion. We rederive expression (12.23) for the time period using this approach. Consider the situation shown in figure (12.11). [Diagram image description: A right triangle is drawn. The length of the hypotenuse is labeled l. The angle of the triangle at the bottom is labeled 0. The hypotenuse is labeled with a point labeled "P" at the end. The side opposite the angle labeled 0 is labeled OQ and the side adjacent to the angle labeled 0 is labeled OP. The angle at point O is labeled with a point labeled O. The particle is labeled with a point labeled "P". A vertical line is drawn from point P to point Q which is on the x-axis. The force of gravity is labeled with a vector labeled mg that acts on point P and points downward. The vector mg is separated into its vertical and horizontal components. mgcosθ is the vertical component of mg that points downwards and mgsinθ is the horizontal component of mg that points to the left.] Suppose the string makes an angle θ with the vertical at time t. The distance of the particle from the equilibrium position along the arc is x = lθ. The speed of the particle at time t is $v=\frac{dx}{dt}$ and the tangential acceleration is $a =\frac{dv}{dt} = \frac{d^2x}{dt^2}$. (i) Forces acting on the particle are (a) the weight mg and (b) the tension T. The component of mg along the tangent to the path is -mgsinθ and that of T is zero. Thus, the total tangential force on the particle is mgsinθ. Using (i) we get or, -mgsinθ = m $\frac{d^2x}{dt^2}$ $\frac{d^2x}{dt^2} = -g sinθ$. (ii) If the amplitude of oscillation is small, sinθ ≈ θ = x/l. Equation (ii) above thus becomes (for small oscillations) $\frac{d^2x}{dt^2} = -g\frac{x}{l}$ or, $\frac{d^2x}{dt^2} = -w²x$ where w² = g/l This equation represents a simple harmonic motion of the particle along the arc of the circle in which it moves. The angular frequency is w = √g/l and the time period is $T = \frac{2\pi}{w}$ $= 2\pi \sqrt{\frac{l}{g}}$ which is same as in equation (12.23). #### Determination of ‘g’ in a Laboratory A simple pendulum provides an easy method to measure the value of ‘g’ in a laboratory. A small spherical ball with a hook is suspended from a clamp through a light thread as shown in figure (12.12). [Diagram image description: A vertical line is drawn with an arrow pointing upwards. At the top of the vertical line is labeled point O and a point labeled P is at the end of the string. The string has a point of attachment at the midpoint labeled A. The particle is labeled with a point labeled P. The particle is at an angle 0 with the vertical line. The force of gravity is labeled with a vector labeled mg that acts on point P and points downward. The vector mg is separated into its vertical and horizontal components labeled mgcosθ and mgsinθ) The lengths AC and BD are measured with slide callipers. The length OA of the thread is measured with a meter scale. The effective length is OP=OA + AP = OA + AC-BD The bob is slightly pulled aside and gently released from rest. The pendulum starts making oscillations. The time for a number of oscillations (say 20 or 50) is measured with a stop watch and the time period is obtained. The value of g is calculated by equation (12.23). The length of the thread is varied and the experiment is repeated a number of times to minimise the effect of random errors. **Example 12.8** In a laboratory experiment with simple pendulum it was found that it took 36 s to complete 20 oscillations when the effective length was kept at 80 cm. Calculate the acceleration due to gravity from these data. **Solution:** The time period of a simple pendulum is given by $T = 2\pi\sqrt{\frac{l}{g}}$ or, $g=\frac{4\pi^2 l}{T^2}$ (i) In the experiment described in the question, the time period is $T = \frac{36 s}{20} =1.8 s$. Thus, by (i), $g = \frac{4\pi^2 × 0.80 m}{(1.8 s)^2} = 9.75 ms^{-2}$. ### 12.10 PHYSICAL PENDULUM Any rigid body suspended from a fixed support constitutes a physical pendulum. A circular ring suspended on a nail in a wall, a heavy metallic rod suspended through a hole in a wall, etc., are examples of a physical pendulum. Figure (12.13) shows a physical pendulum. A rigid body is suspended through a hole at its centre of mass C vertically below point O. When the centre of mass is vertically below point O the body may remain at rest. We call this position the equilibrium position. When the body is pulled aside and released, it executes oscillations. [Diagram image description: A rigid body is drawn. The body is suspended from point by a string. Point A, which is the equilibrium position of the body is directly below point O. Point C, represents the center of mass of the object that is suspended on the string. Points B and D are located on the body of the object. The particle is labeled with a point labeled "P". A vertical line is drawn from point P to point Q which is on the x-axis. The force of gravity is labeled with a vector labeled mg that acts on point C and points downward.] The body rotates about a horizontal axis through O and perpendicular to the plane of motion. Let this axis be OA. Suppose the angular displacement of the body is θ at time t. The line OC makes an angle θ with the vertical at this instant. Forces on the body are (a) the weight mg and (b) the contact force V by the support at O. The torque of V about OA is zero as the force V acts through the point O. The torque of mg has magnitude |Γ| = mg (OD) = mg (OC) sin θ = mgl sinθ where l=OC is the separation between the point of suspension and the centre of mass. This torque tries to bring the body back towards θ = 0. Thus, we can write Γ = -mglsinθ. If the moment of inertia of the body about OA is I, the angular acceleration becomes $α = \frac{-mgl}{I}sinθ$. (i) We see that the angular acceleration is not proportional to the angular displacement and the motion is not strictly simple harmonic. However, for small displacements sinθ ≈ θ so that equation (i) becomes $α = -\frac{mgl}{I}θ$ or, $α = -w²θ$ where $w² = \frac{mgl}{I}$ Thus, for small oscillations, the motion is nearly simple harmonic. The time period is $T = \frac{2\pi}{w}$ $= 2\pi \sqrt{\frac{I}{mgl}}$. (12.24) **Example 12.9** A uniform rod of length 1.00 m is suspended through an end and is set into oscillation with small amplitude under gravity. Find the time period of oscillation. **Solution:** For small amplitude the angular motion is nearly simple harmonic and the time period is given by $Τ = 2π\sqrt{\frac{I}{mgl}}$ $= 2π\sqrt{\frac{mL^2/3}{mgL/2}}$ $= 2π \sqrt{\frac{2L}{3g}}$ $= 2π\sqrt{\frac{2 × 1.00 m}{3 × 9.80 m s^{-2}}}$ = 1.64 s. ### 12.10 TORSIONAL PENDULUM In torsional pendulum, an extended body is suspended by a light thread or a wire. The body is rotated through an angle about the wire as the axis of rotation (figure 12.14). [Diagram image description: A rigid body is attached to a string that is vertical. The end of the string is connected to a support, the string is twisted, and the angle between the string and the horizontal is labeled θ. The body rotates clockwise.] The wire remains vertical during this motion but a