NAEST Exp2

Roll No: 21MP038 Name: Mohammad Saif Khan Name of College and Address: Deshbandhu College, University of Delhi. Kalkaji Main Rd, Block H, Kalkaji, New Delhi, Delhi 110019

Experiment Name and No. : OSCILLATIONS, EXPT-2

Acknowledgements : A special thanks to my father, who took time from his busy schedule to help me with videography and photography of the experiments along with helping me save time in setting up the experimental setup.

Introduction In the following experiments, the behaviour of complex oscillatory systems is studied. From their time periods to the general trajectories of different kinds of systems is studied. Also their dependence on properties of oscillating objects is also taken into consideration.

Part A : Studying the effect of changing of mass on a coupled pendulum.

Part B : Observing the oscillation of a multi-frequency pendulum. These systems are much more complex compared to an ordinary pendulum, this gives much more insight on the phenomenon of oscillations and harmonic motion.

PART A Study of behaviour of a Coupled Pendulum, and analyze the dependence of time period on mass.

Materials: ● A drawstring is used as the uppermost horizontal string attached to two sides of door space with cellotape. ● Two erasers are used as bobs, attached with strings with the help of knitting needles. ● Erasers are used as we know their mass beforehand from packaging and it is very easy to make a pendulum with the use of knitting needles. Pic 1 ● The weighing pan is made out of thin mountboard. ● The ‘bob’ and pan are joined and coupled using regular cloth strings. ● For variable weights, a small moisturizer container is used with known manufactured volume and mass.

Pic 2 ● Water is put inside the container using a standard teaspoon.

Pic 3 - Setup

Theoretical Analysis: We have with ourselves a String Coupled Pendulum [1], with a variable middleweight as shown in the figure.

Fig 1

Fig 2 Here we see that tension in the middle string causes it to act equivalent to a spring coupled system with the string causing a restoring force. The middleweight has a varying mass, this allows us to change the tension in the middle string and essentially have a spring where we can vary the spring constant 𝑘 In the setup there are primarily two forces responsible for the motion of the pendulum :

1.) When any one of the pendulum is displaced by and angle θ, perpendicular to the plane of the setup, It suffers restoring gravitational torque : τ𝑔 =− 𝐹𝐿𝑠𝑖𝑛(θ), 𝐹 = (𝑚𝑔) 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑖𝑛𝑔 𝑠𝑚𝑎𝑙𝑙 𝑎𝑛𝑔𝑙𝑒𝑠, 𝑠𝑖𝑛(θ) ≈ θ 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, τ𝑔 =− 𝑚𝑔𝐿θ 𝑚 is the mass of one bob, 𝑔 is acceleration due to gravity, 𝐿 is the perpendicular distance from the rotational axis, i.e. length of string. 2.) Let the two pendula be displaced by amounts, θ1 and θ2. The coupling torque is caused by the string, since it is an equivalent spring we can define it in terms of : ● spring constant 𝑘, ● distance between bob and point of coupling 𝑙 and ● the difference between the angular displacement of two bobs (Δθ) 2

τ𝑓 =− 𝑘𝑙 𝑠𝑖𝑛Δθ 𝑠𝑖𝑛(θ) ≈ θ 2

τ𝑓 =− 𝑘𝑙 Δθ The mid string is already slightly in tension which causes pendulums to be at rest at an angle α and − α respectively relative to the vertical position when the pendulums are at rest. The torque produced by this cancels itself out when added for both the bobs. So the effect will be ignored. We have : 𝑇𝑜𝑟𝑞𝑢𝑒 𝑜𝑛 𝑏𝑜𝑏 1, τ1 = τ𝑔,1 + τ𝑓,1

𝑇𝑜𝑟𝑞𝑢𝑒 𝑜𝑛 𝑏𝑜𝑏 2, τ2 = τ𝑔,2 + τ𝑓,2 2

𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, τ1 =

− 𝑚𝑔𝐿θ1 − 𝑘𝑙 (θ1 − θ2) 2

τ2 =

− 𝑚𝑔𝐿θ2 − 𝑘𝑙 (θ2 − θ1)

Torque can also be defined as : 2

2

τ = 𝐼 · (𝑑 θ/𝑑𝑡 ), 𝐼 𝑖𝑠 𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝐼𝑛𝑒𝑟𝑡𝑖𝑎 We have : 2

2

2

2

𝑑 (θ1)/𝑑𝑡 = 𝑑 (θ2)/𝑑𝑡 =

2

− (𝑔𝐿/𝐼)θ1 + (𝑘𝑙 /𝐼)(θ2 − θ1) 2

− (𝑔𝐿/𝐼)θ2 − (𝑘𝑙 /𝐼)(θ2 − θ1)

Adding the above equations we get : 2

2

𝑑 (θ2 + θ1)/𝑑𝑡 = 2

2

𝑑 (θ2 − θ1)/𝑑𝑡 =

− (𝑔𝐿/𝐼)(θ2 + θ1) 2

− [(𝑔𝐿/𝐼) + 2(𝑘𝑙 /𝐼)](θ2 − θ1)

Writing it in the form of angular frequency : 2

2

2

2

𝑑 (θ2 + θ1)/𝑑𝑡 = 𝑑 (θ2 − θ1)/𝑑𝑡 =

2

− ω𝑎 (θ2 + θ1) 2

− ω𝑠 (θ2 − θ1)

These equations are nothing but differential equations of Simple Harmonic Motion, giving us the general solution to be :

(θ2 + θ1) = 2𝐴 · 𝑐𝑜𝑠 (ω𝑎𝑡 ) (θ2 − θ1) = 2𝐵 · 𝑐𝑜𝑠 (ω𝑠𝑡 + δ ) Or θ1 = 𝐴 · 𝑐𝑜𝑠 (ω𝑎𝑡 ) − 𝐵 · 𝑐𝑜𝑠(ω𝑠𝑡 + δ ) θ2 = 𝐴 · 𝑐𝑜𝑠 (ω𝑎𝑡 ) + 𝐵 · 𝑐𝑜𝑠(ω𝑠𝑡 + δ ) 𝐴, 𝐵 𝑎𝑟𝑒 𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒𝑠. δ 𝑖𝑠 𝑡ℎ𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑝ℎ𝑎𝑠𝑒 ω𝑎 = 𝑚𝑔𝐿/𝐼, ω𝑠 =

2

(𝑔𝐿 + 2𝑘𝑙 )/𝐼

From our general solution we can study 3 special cases and their Time Periods : Case 1 : When the two bobs are displaced together such that, θ1 = θ2 = θ. This is the In Phase Condition where : θ = 𝐴 · 𝑐𝑜𝑠 (ω𝑎𝑡 ) 𝑇𝑖𝑚𝑒 𝑃𝑒𝑟𝑖𝑜𝑑, 𝑇1 = 2π/ω𝑎 𝑇1 = 2π 𝐼/𝑚𝑔𝐿 We observe that there is no involvement of the midweight due to it suffering no compressions. This is called symmetric oscillation. The equation of time period is of the form : 𝑦 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Prediction 1 : Time Period T1 should be constant

Case 2 : When the two bobs are displaced together such that, θ = θ2 = − θ1. This is the Out of Phase Condition where : θ = 𝐵 · 𝑐𝑜𝑠 (ω𝑠𝑡 + δ ) 𝑇𝑖𝑚𝑒 𝑃𝑒𝑟𝑖𝑜𝑑, 𝑇1 = 2π/ω𝑠 2

𝑇1 = 2π (𝐼/𝑚𝑔𝐿) + (𝐼/2𝑘𝑙 ) This is called an asymmetric oscillation. The equation of time period is of the form : 𝑦 ∝ 1/ (𝑐 + 𝑥), 𝑐 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Hence, 𝑓𝑜𝑟 𝑥 ∈ (− 𝑐, + ∞), 𝑦 ∈ (+ ∞, 0) Prediction 2 : Time Period T2 should be decreasing w.r.t mass of midweight. Case 3 : When the two bobs are displaced together such that, θ1 = 0 and θ2 = θ. This is the Beats Condition where, when one pendulum oscillates and transfers its energy to the other pendulum. Also the amplitude is 𝐴 = 𝐵 = 𝑍 and the phase difference is 0. Eventually, its oscillations die down after some time and the other pendulum picks up oscillations gradually. This process happens in reverse after a certain time period. We have from general solutions :

θ1 = 𝑍 · (𝑐𝑜𝑠 (ω𝑎𝑡 ) − 𝑐𝑜𝑠(ω𝑠𝑡)) θ2 = 𝑍 · (𝑐𝑜𝑠 (ω𝑎𝑡 ) + 𝑐𝑜𝑠(ω𝑠𝑡)) This gives : θ1 = 2𝑍 · 𝑠𝑖𝑛[((ω𝑠 − ω𝑎)/2)𝑡] · 𝑠𝑖𝑛[((ω𝑠 + ω𝑎)/2)𝑡] θ2 = 2𝑍 · 𝑐𝑜𝑠[((ω𝑠 − ω𝑎)/2)𝑡] · 𝑐𝑜𝑠[((ω𝑠 + ω𝑎)/2)𝑡] Here we can define : ω𝑏 = (ω𝑠 − ω𝑎)/2, 𝑡ℎ𝑖𝑠 𝑖𝑠 𝑡ℎ𝑒 𝐵𝑒𝑎𝑡 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦

ω𝑎𝑣 = (ω𝑠 + ω𝑎)/2, 𝑡ℎ𝑖𝑠 𝑖𝑠 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 For weak coupling as it is in our setup : ω𝑎𝑣 >> ω𝑏 Thus we will observe Harmonic Motion but with periodically changing Amplitude. There is also a phase difference of π/2 between θ1 and θ2 .

Fig 3 Time period T3 is the time difference between two instances when the same pendulum is at rest and is called Beat Period. Since it is the lower frequency : 𝑇3 = 2π/(ω𝑠 − ω𝑎) This an equation of the form : 𝑦 ∝ 1/( 𝑐1 + 𝑥 −

𝑐2)

Therefore : 𝑓𝑜𝑟 𝑥 ∈ (− 𝑐, + ∞) 𝑎𝑛𝑑 𝑐1 + 𝑥 ≠ 𝑐2, 𝑦 ∈ (+ ∞, 0) Prediction 3 : Time Period T3 should be asymptotically decreasing w.r.t mass of midweight. Prediction 4 : We observe the different time periods in previous three predictions : 𝑇1 = 2π/ω𝑎 𝑇2 = 2π/ω𝑠 𝑇3 = 2π/(ω𝑠 − ω𝑎) Therefore : 1/𝑇3 = 1/𝑇2 − 1/𝑇1

Methodology: To study the effects of a coupled pendulum, the two bobs have been kept far apart to avoid collision with each other or with the weighing pan. The oscillations are done perpendicular to the plane. The weight of the strings and the pan is considered to be negligible. The container is put firmly on top of the pan, this acts as a water container that we can use to vary the weight on the pan hence influencing the time periods. The water in the container is added using a standard teaspoon which gives us a standard measurement of volume of water at 5 𝑚𝐿. Multiplying this with the density of water at room temperature, ρ = 0. 997 𝑔/𝑚𝐿 𝑎𝑡 25°𝐶, we can to a very good approximation measure the mass of water added. The oscillations are kept small enough for the motion to be approximately harmonic and effects of damping have been ignored. The three different time periods are measured using an electronic stopwatch.

T1 : In Phase In this case the bobs are displaced by the same amount in the same direction, time for 10 oscillations is measured and divided by 10 for the period.

T2 : Out of Phase In this case the bobs are displaced by the same amount in the opposite direction, time for 10 oscillations for one of the bobs is measured and divided by 10 for the period.

T3 : Beats In this case keeping one of the pendulum stationary and displacing the other through a distance, we observe that the oscillations of the displaced pendulum die down after some time and the stationary pendulum picks up oscillations gradually. After some time this process repeats in reverse. The time period is the time between two stationary positions (zero oscillation states) of any one of the two bobs.

Numerical Analysis: 𝐿𝑒𝑎𝑠𝑡 𝐶𝑜𝑢𝑛𝑡 𝑜𝑓 𝑆𝑡𝑜𝑝𝑤𝑎𝑡𝑐ℎ = 0. 1 𝑠𝑒𝑐 𝑀𝑎𝑠𝑠 𝑜𝑓 𝐶𝑜𝑛𝑡𝑎𝑖𝑛𝑒𝑟, 𝑊 = 17𝑔 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑜𝑛𝑒 𝐸𝑟𝑎𝑠𝑒𝑟, 𝑚 = 𝑔 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟, ρ = 0. 997 𝑔/𝑚𝐿 𝑎𝑡 25°𝐶

Volume of Water V (mL)

Mass of Water m’ (g)

Total In Out of Beat Mass on Phase Phase Period Pan Period Period T3 (s) M=W+m’ T1 (s) T2 (s) (g)

0

0

17

10

9.970

26.970

20

19.940

36.940

35

24.925

41.925

50

49.850

66.850 Table A1

Graphs : 1. 𝑇1, 𝑇2, 𝑇3 vs 𝑀 Scale : X axis - 1 cm = units Y axis - 1 cm = units

Graph A

Conclusions : We observe from Table A that : ● T1 remains constant in the seconds range, the uncertainty in the millisecond range can be attributed to human reaction time. This confirms our Prediction 1. ● T2 ● T3 From T1, T2, T3 : 1/T1

1/T2

1/T2 - 1/T1

1/T3

%Error

Table A2 Observing Table A2, we can observe that within a reasonable margin of error : 1/𝑇3 = 1/𝑇2 − 1/𝑇1 This confirms our Prediction 4

PART B Study the shadow of a thin obstacle placed on a mirror.

Materials : ● A drawstring is used as the horizontal string as shown in Fig a. ● A small 250mL bottle is filled with chuuna with a hole in the bottom sealed with cello tape.

Pic 4 ● Another string is used to create a pendulum off the bottle. ● A cardboard piece acts as a screen for the sand patterns.

Pic 5 - Setup

Theoretical Analysis :

Fig 4

Fig 5

Our discussion of such a system is based on oscillations, that unlike simple oscillations, present in 2D with an 𝑥 and 𝑦 components.[2] Let the 𝑥 axis be in the plane of the pendulum M and the 𝑦 axis be perpendicular to plane M. Therefore, the angle at which we displace our bob w.r.t to plane M θ. What we have with ourselves is essentially a Spherical Pendulum with θ acting as the Azimuthal Angle :

𝑥 = 𝑙𝑐𝑜𝑠(θ) · 𝑠𝑖𝑛(ω𝑡) 𝑦 = 𝑙𝑠𝑖𝑛(θ) · 𝑠𝑖𝑛(ω𝑡) 𝑧 = 𝑙𝑐𝑜𝑠(ω𝑡) Where : 𝑙 𝑖𝑠 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑡𝑟𝑖𝑛𝑔 (L2 in Fig4) ω 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 We ignore the vertical displacement of the bottle for the time being, as we are concerned with the 2D projection of the trajectory of the bottle on the XY Plane (on the floor). From [Fig 5] and the equations of Spherical Pendulum we have: Prediction 1 : When θ = 0°, i.e. the oscillations are in the plane, or : Y amplitude is 0

Prediction 2 : When θ = 90°, i.e. the oscillations are perpendicular to the plane, or : X amplitude is 0 We can also observe from the top view of the setup :

Fig 6

The force on bottle 𝐹𝑦 causes the top sting to move with a displacement of Δ𝑑. The force 𝐹𝑥 also causes a slight change of position of the top string as seen from the front.

Fig 7 The effects don’t come into play in previous predictions as one of the amplitudes is always 0. This adds another layer of oscillation to the system by slightly moving the fixed point of the system. (The effects are slightly exaggerated for explanation purposes.) From this we have :

Prediction 3 : When 0 < θ < 90°, 𝑥 = 𝑐1 · 𝑠𝑖𝑛(ω𝑡) 𝑦 = 𝑐2 · 𝑠𝑖𝑛(ω𝑡 + δ) Solving these equations we get : 2

2

2

(𝑥/𝑐1) + (𝑦/𝑐2) − 2𝑥𝑦𝑐𝑜𝑠(δ)/𝑐1𝑐2 = 𝑠𝑖𝑛 (δ) This is a general equation of an ellipse But due to the movement of the top string, we will observe a varying phase difference. 𝑥 = 𝑐1 · 𝑠𝑖𝑛(ω1𝑡+ δ1) 𝑦 = 𝑐2 · 𝑠𝑖𝑛(ω2𝑡 + δ2) ω1, ω2, δ1 and δ2 change due to the movement of the top string. Therefore : We should get a flower-like pattern made up of smaller constituent non constant elliptical paths

Methodology : In the experiment a plane of oscillation 𝑀 is decided and the bottle is moved at various angles with respect to the plane. Case I : Oscillations in the plane are studied by displacing the bottle within the plane 𝑀. Case II : Oscillations perpendicular to the plane are studied by displacing the bottle at 90° to the plane 𝑀. Case III : Oscillations are studied when the bottle was displaced at an acute angle with respect to the plane 𝑀. The trajectories of these oscillations are studied and compared. Then we displace the bottle at an angle of 30° w.r.t the plane 𝑀, and the cello tape at the bottom is removed. The sand patterns are observed. This is repeated for 45° and 60° w.r.t plane 𝑀.

Visual Analysis : For Cases I and II, it was observed that the 2D projection of the bottle was a straight line, parallel to the x and y axis respectively. This is in accordance with our Prediction 1 and 2. (Here the y axis is shown)

Pic 6 For Case III, we see the bottle moving on a path made up of small elliptical paths, making a flower-like pattern. Also letting the sand fall we can also show the projection of the trajectory on XY plane :

1.)

θ = 30°

Pic 7 2.)

θ = 45°

Pic 8

3.)

θ = 60°

Pic 9

This is accordance with our Prediction 3 Theoretical Analysis matches with the Visual Analysis

Exploration Part B : Observing the effect of no top string and verifying our predictions. In our discussion of the path the bottle takes, we concluded that the oscillatory motion of the top string, due to it not being a rigid body and the forces on the bottle move the top string. According to our Theoretical analysis, if the top string doesn’t move it would not give the flower-like pattern and will be governed by the equations : 𝑥 = 𝑐1 · 𝑠𝑖𝑛(ω𝑡) 𝑦 = 𝑐2 · 𝑠𝑖𝑛(ω𝑡 + δ) Solving these equations we get : 2

2

2

(𝑥/𝑐1) + (𝑦/𝑐2) − 2𝑥𝑦𝑐𝑜𝑠(δ)/𝑐1𝑐2 = 𝑠𝑖𝑛 (δ) This is nothing but the equation of a general ellipse. In our extra exploration we remove the contribution of the top string by holding the pendulum with our hands.

Pic 10 This results in the top string being tensionless. This gives a generic spherical pendulum. If the azimuthal angle (angle made with plane M) is kept constant, the path should follow the above equations and give us the ellipse.

We oscillate our bottle just like before with an acute azimuthal angle.

Pic 11

We remove the cellotape to observe the trajectory and it is exactly as we expect. An ellipse is formed :

Pic 12 This is in accordance with our prediction and reconfirms the matching of theoretical and visual analysis.

References : [1] Young-Ki Cho, “Teaching the Physics of a String-Coupled Pendulum Oscillator: Not Just for Seniors Anymore”. The Physics Teacher 50, 417 (2012); https://doi.org/10.1119/1.4752047. [2] A.P. French, Vibration and Waves (MIT Introductory Lectures), Chapter 2 “The superposition of periodic motion”.