Lab Experiments PDF

Experiment 1: Simple Pendulum Student’s Name ID number 10 Objective: - To determine the acceleration due to gravity, g. Theory: Periodic motion, the repeating motion of an object in which it continues to return to a given po...

Experiment 1: Simple Pendulum Student’s Name ID number 10 Objective: - To determine the acceleration due to gravity, g. Theory: Periodic motion, the repeating motion of an object in which it continues to return to a given position after a fixed time interval. The repetitive movements of such an object are called oscillations. The simple pendulum is a mechanical system that exhibits periodic motion. It consists of a particle-like bob of mass 𝒎 suspended by a light string of length 𝑳 that is fixed at the upper end as shown in the figure. The motion occurs in the vertical plane and is driven by the gravitational force. The period (𝐓) of a pendulum is the time interval required for the particle to go through one complete oscillation of its motion. The frequency (𝐟) of a pendulum represents the number of oscillations the particle undergoes per unit time interval. The frequency (𝐟) and the period (𝐓) of a pendulum in terms of number of oscillations are given by: 𝐍 𝐟= (𝟏) 𝐭 𝐭 𝐓= (𝟐) 𝐍 𝟏 𝐟= (𝟑) 𝐓 where 𝐍 = 𝐧𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐜𝐨𝐦𝐩𝐥𝐞𝐭𝐞 𝐨𝐬𝐜𝐢𝐥𝐥𝐚𝐭𝐢𝐨𝐧𝐬 and 𝐭 = 𝐭𝐨𝐭𝐚𝐥 𝐭𝐢𝐦𝐞 𝐢𝐧 𝐬𝐞𝐜𝐨𝐧𝐝𝐬. The frequency (𝐟) and the period (𝐓) of a pendulum of length 𝐋, with free-fall acceleration g, are given by: 𝟏 𝐠 𝐟= √ (𝟒) 𝟐𝛑 𝐋 1  𝐋 𝐓 = 𝟐𝛑√ (𝟓) 𝐠 with 𝐠 = 𝟗. 𝟖 m/s2. One can square the period to obtain: 𝟐 𝟒𝛑𝟐 𝐓 = 𝐋 (𝟔) 𝐠 The acceleration due to gravity can be calculated from the slope of the line resulting from the above relationship: 𝟒𝛑𝟐 𝐠 𝐞𝐱𝐩 = (𝟕) 𝐬𝐥𝐨𝐩𝐞 Data: 𝐭 𝟒𝛑𝟐 𝐋 L (m) 𝐭 𝟐𝟎 ( s ) 𝐓 = 𝟐𝟎 (𝒔) 𝟐 𝐓 (s ) 2 𝐠= (𝐦/𝐬 𝟐 ) 𝟐𝟎 𝐓𝟐 0.3 22.2 …………… …………… ………….…………… 0.4 25.4 …………… …………… ………….…………… 0.5 28.0 …………… …………… ………….…………… 0.6 30.8 …………… …………… ………….…………… 0.7 33.6 …………… …………… ………….…………… 0.8 35.8 …………… …………… ………….…………… 1. Arithmetic method: i. Complete the table above ii. Calculate the average value, g av 𝐠 𝟏 + 𝐠 𝟐 + 𝐠 𝟑 + 𝐠 𝟒 + 𝐠 𝟓 + 𝐠 𝟔 …. + …. + ⋯ + ⋯ + … +⋯ 𝐠 𝐚𝐯 = = =⋯ 𝐦/𝐬 𝟐 𝟔 𝟔 2. Graphical method: i. Plot a graph of 𝐓 𝟐 (y-axis) against L (x-axis). Use the graph paper on page 3 (Display Titles and units of x and y-axis) ii. Determine the slope of the graph 𝒚𝟐 − 𝒚𝟏 … − … 𝐬𝐥𝐨𝐩𝐞 = = =⋯ 𝐬 𝟐 /𝒎 𝒙𝟐 − 𝒙𝟏 … − … 2  iii. Calculate the experimental value, g exp , using the formula: 𝟒𝛑𝟐 𝟒𝛑𝟐 𝐠 𝐞𝐱𝐩 = = = ….. 𝐦/𝐬 𝟐 𝐬𝐥𝐨𝐩𝐞 … iv. Calculate the error percent between the experimental value of g (g exp ) and its standard value (g = 9.8 m. s −2 ) |𝐠 𝐞𝐱𝐩 − 𝐠| | … − 𝟗. 𝟖| 𝐞𝐫𝐫𝐨𝐫 % = × 𝟏𝟎𝟎 = × 𝟏𝟎𝟎 = ⋯ % 𝐠 𝟗. 𝟖 (Please use the graph paper below) 4.0 3.6 3.2 2.8 2.4 2 1.6 1.2 0.8 0.4 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 3 Check: 1. In the opposite figure. The pendulum makes a complete oscillation when it moves from …………….. 2. In the opposite figure. If the time taken by the pendulum to move from X to Z is 0.8 s, calculate: a) The period (𝐓). b) The frequency (𝐟). c) The number of complete oscillations (𝐍) in 16 s. d) The time (𝐭) required to make 50 oscillations. 3. In the opposite figure. If the pendulum takes 1.0 s to move from X to Y, so its frequency (𝐟) equals ………….. 4. If a pendulum makes 30 complete oscillations per one minute, what is the frequency (𝐟) and the period (T) of the pendulum motion? 5. A simple pendulum of length 𝐋 = 𝟎. 𝟒 𝐦, what are the frequency (𝐟) and period (T) of its motion? 4  Experiment 2: Speed of Sound Student’s Name ID number 10 Objective: To determine the speed of sound in air. Theory: A sound wave is a longitudinal wave in which the wave oscillates along the direction of propagation. For a traveling sound wave of speed (𝐯), frequency (𝐟) and wavelength (𝝀), in a resonance tube (closed at one end), the distance between a node and an anti-node is λ/ 4. It is found experimentally that the resonance occurs when the length of the air column at the resonance (𝐋) is nearly equal to an odd number of λ /4 i.e. 𝐧𝛌 𝐋= (𝐧 = 𝟏, 𝟑, 𝟓, … ) (𝟏) 𝟒 In other words; 𝟒𝐋 𝝀= (𝐧 = 𝟏, 𝟑, 𝟓, … ) (𝟐) 𝐧 Since the speed of a wave reads; 𝐯 = 𝐟𝛌 then, 𝐯 𝐋=𝐧 (𝐧 = 𝟏, 𝟑, 𝟓, … ) (𝟑) 𝟒𝐟 The resonance frequencies, 𝐟𝐧 , of a tube (air column) depend on its length 𝐋; 𝐯 𝐟𝐧 = 𝐧 (𝐧 = 𝟏, 𝟑, 𝟓, … ) (𝟒) 𝟒𝐋 The lowest frequency of (𝐧 = 𝟏) is called the fundamental frequency and higher frequencies are called overtones. For the fundamental frequency, from equation (3) the following relationship holds: 5  𝐯 𝟏 𝐋= 𝟒 𝐟 The speed of sound can be calculated from the slope of the line resulting from the above relationship: 𝐯𝐞𝐱𝐩 = 𝟒 × 𝐬𝐥𝐨𝐩𝐞 According to the theory, the speed of sound in air depends upon the temperature of the air through the following relationship: 𝐯𝐭𝐡 = 𝐯𝟎 + 𝟎. 𝟔 𝐓 Here 𝐯𝟎 = 𝟑𝟑𝟏 𝒎/𝒔 is the speed of sound at 𝟎𝐨 𝐂 , and 𝐓 is the temperature in the Lab (say; 𝐓 = 𝟐𝟎𝐨 𝐂 ). Data: 𝟏 𝐟 (Hz) (× 𝟏𝟎−𝟑 𝐬) 𝐋 ( × 𝟏𝟎−𝟐 𝒎) 𝐯= 𝟒𝐟𝐋 (𝐦/𝐬) 𝐟 500 17 417 20.4 357 23.8 312 27.2 277 30.6 1. Arithmetic method: Calculate the average value, vav ? 𝐯𝟏 + 𝐯𝟐 + 𝐯𝟑 + 𝐯𝟒 + 𝐯𝟓 … + … + … + … + … 𝐯𝐚𝐯 = = = … 𝐦/𝐬 𝟓 𝟓 2. Graphical method: i. Plot 1/f on the x-axis and L on the y-axis. Use the graph paper on page 7 (Display Titles and units of x and y-axis) ii. Determine the slope of the graph: 𝒚𝟐 − 𝒚𝟏 … − … 𝐬𝐥𝐨𝐩𝐞 = = = … 𝐦/𝐬 𝒙𝟐 − 𝒙𝟏 … − … iii. Calculate the experimental value, vexp ? 𝐯𝐞𝐱𝐩 = 𝟒 × 𝐬𝐥𝐨𝐩𝐞 = 𝟒 × …. = … 𝐦/𝐬 6 3. The percentage error: 𝐯𝐭𝐡 = 𝐯𝟎 + 𝟎. 𝟔 𝐓 = ⋯ + 𝟎. 𝟔 × … = ⋯ 𝐦/𝐬 |𝐯𝐞𝐱𝐩 − 𝐯𝐭𝐡 | |…. −... | 𝐞𝐫𝐫𝐨𝐫 % = × 𝟏𝟎𝟎 = × 𝟏𝟎𝟎 = ⋯ % 𝐯𝐭𝐡 … (Please use the graph paper below) 7 Check: 1. Consider a tube of length 𝐋 = 𝟎. 𝟑 𝒎 that is closed at one end. What is the wavelength of the lowest tone produced by this tube? 2. Consider a tube of length 𝐋 = 𝟏. 𝟎 𝐦 that is closed at one end and the travelling sound wave has a speed 𝐯 = 𝟖𝟎 𝐦/𝐬. (a) What is the fundamental frequency produced by this tube? (b) What is the lowest overtone frequency produced by this tube? 3. What is the speed of a travelling sound wave, 𝐯𝐭𝐡 , in air at a room of temperature 𝐓 = 𝟐𝟎𝐨 𝐂 ? (𝐯𝟎 = 𝟑𝟑𝟏 𝒎/𝒔 is the speed of sound at 𝟎𝐨 𝐂) 8  Experiment 3: Lenses Student’s Name ID number 10 Objectives: (1) To determine the focal length (𝐟) of a converging lens. (2) To determine the power (𝐏) of a converging lens. Theory: A lens is a transparent object that uses refraction of light rays at curved surfaces to form an image. Lenses are usually used to converge or diverge the light rays. A converging lens refract the light rays toward its axis. A diverging lens refract the light rays outward from its axis. The focal length of a lens (𝐟) is the distance from the center of the lens to the focal point. It is positive for the converging lens, but negative for the diverging lens. For thin length, the relation between the focal, object and image lengths is: 𝟏 𝟏 𝟏 = + (𝟏) 𝐟 𝐬 𝐬′ where 𝐬 is the object distance and 𝐬′ is the image distance. Then, the focal length reads: 𝐬 × 𝐬′ 𝐟= (𝟐) 𝐬 + 𝐬′ Power of a lens is a measure of how much the lens bends rays of light. A high- power lens bends light rays through a large angle (𝜃), leading to a short focal length. A low-power lens bends the rays through a small angle (𝜃), leading to a long focal length. 9  The power of a lens (𝐏) is equal to the reciprocal of its focal length; 𝟏 𝐏= (𝐃 ≡ 𝐦−𝟏 ) (𝟑) 𝐟 The SI unit of power of a lens is “Diopter (D)”. If the focal length, 𝐟 (𝐜𝐦), is measured in 𝐜𝐦 , the power of the lens is given by: 𝟏𝟎𝟎 𝐏= (𝐜𝐦−𝟏 ) (𝟒) 𝐟 Remark: 𝐃 ≠ 𝐜𝐦−𝟏. Data: 𝟏 𝟏 𝐬 × 𝐬′ 𝐬 (cm) 𝐬′ (𝐜𝐦) (𝐜𝐦−𝟏 ) (𝐜𝐦−𝟏 ) 𝐟= (𝐜𝐦) 𝐬 𝐬′ 𝐬 + 𝐬′ 10 66.5 12.5 33 16.5 22 25 16.5 50 13.3 1. Arithmetic method: i. Calculate the average value, fav ,? 𝐟𝟏 + 𝐟𝟐 + 𝐟𝟑 + 𝐟𝟒 + 𝐟𝟓 … + … + … + … + … 𝐟𝐚𝐯 = = = … 𝐜𝐦 𝟓 𝟓 ii. Calculate the power of the lens? 𝟏𝟎𝟎 𝟏𝟎𝟎 𝐏= = =⋯ 𝐜𝐦−𝟏 𝐟 … 10 2. Graphical method: i. Plot 1/s on the x-axis and 1/s′ on the y-axis. Use the graph paper on page 11 (Display Titles and units of x and y-axis) ii. Determine the 𝑥 and 𝑦 coordinates? 𝒙 − 𝐜𝐨𝐨𝐫𝐝𝐢𝐧𝐚𝐭𝐞: 𝐎𝐌 = … 𝐜𝐦−𝟏 𝒚 − 𝐜𝐨𝐨𝐫𝐝𝐢𝐧𝐚𝐭𝐞: 𝐎𝐊 = … 𝐜𝐦−𝟏 iii. Calculate the experimental value, fexp ,? 𝟐 𝐟𝐞𝐱𝐩 = =⋯ 𝐜𝐦 𝐎𝐌 + 𝐎𝐊 iv. Calculate the power of the lens, P,? 𝟏𝟎𝟎 𝐏= =⋯ 𝐜𝐦−𝟏 𝐟𝐞𝐱𝐩 (Please use the graph paper below) 11  Check: 1- The Power of a lens is −𝟐. 𝟓 Diopter. (a) What is the focal length of the lens? (b) Is it a converging or a diverging lens? 2- Which of the lenses with focal length 10 cm, 20 cm, 25 cm and 50 cm has maximum power? a) 10 cm b) 20 cm c) 25 cm d) 50 cm 3- A focal length of a lens is 10 cm. What is the power of this lens in diopter? a) 0.1 D b) 10 D c) 1 D d) 100 D 4- The SI unit of the power of lens is a) Dioptre b) cm−1 c) Watt d) m 12  Experiment 4: Viscosity of Liquids Student’s Name ID number 10 Objective: To measure the dynamic viscosity of a liquid by using a falling sphere (Stokes method). Theory: Viscosity is a general property of fluids which determine their tendency to oppose their flow. It occurs as a result of contact fluid layers with each other and can be valued as dynamic (absolute) or kinematic (ratio of dynamic viscosity to density). The dynamic viscosity is usually denoted by the Greek symbol 𝛍 (mu) and is defined as the ratio of the shear stress to the shear rate of the fluid. Shear stress, 𝜏 gradient, 𝜕𝑢⁄𝜕𝑦 Figure 1: Friction between fluid and boundaries causes shear stress at a specific gradient The shear stress (denoted 𝛕 :Greck letter tau) is dependent on the fluid’s resistance force 𝐅 to flow over the area 𝐀 of the plate while the shear rate is equivalent to the velocity gradient 𝐝𝐮⁄𝐝𝐲 within the fluid (where this is the derivative of the velocity with respect to the distance 𝐲): 𝐅 𝐬𝐡𝐞𝐚𝐫 𝐬𝐭𝐫𝐞𝐬𝐬 𝛕 𝐀 𝛍= = = (𝟏) 𝐬𝐡𝐞𝐚𝐫 𝐫𝐚𝐭𝐞 𝐝𝐮⁄ 𝐝𝐮⁄ 𝐝𝐲 𝐝𝐲 From the above equation it can be seen that the SI unit for dynamic viscosity is Pa. s or N. s. m−2 or Kg. m−1. s −1 : 13  𝟏𝐏𝐚. 𝐬 = 𝟏𝐍. 𝐬. 𝐦−𝟐 = 𝐤𝐠. 𝐦−𝟏. 𝐬 −𝟏 (𝟐) The viscosity of a liquid, which is extensively governed by the temperature of the medium, can be measured using a viscometer. A viscometer consists of a graduated glass cylinder filled with the liquid. Small steel spheres are dropped into the liquid from the top. After falling sufficiently, the spheres acquire a constant speed, 𝐯𝐟 , which is called the terminal speed. When a solid sphere falls in a liquid, a viscous drag force 𝐅𝐝 will be exerted on the sphere. According to Stokes’s law, the drag force is proportional to the viscosity 𝛍 of the liquid, the radius 𝐫 of the sphere, and the speed 𝐯 of the sphere as: 𝐅𝐝 = 𝟔𝛑𝛍𝐫𝐯 (𝟑) Simultaneously it experiences a gravitational force, 𝐖, and a buoyant force, 𝐅𝐛 as shown in Figure.2: 𝟒 𝟑 𝐖 = 𝐦𝐠 = 𝐕𝛒𝐠 = 𝛑𝐫 𝛒𝐠 (𝟒) 𝟑 𝟒 𝟑 𝐅𝐛 = 𝐕𝛒𝟎 𝐠 = 𝛑𝐫 𝛒𝟎 𝐠 (𝟓) 𝟑 where 𝐕 is the volume of the sphere ball, 𝛒 is the density of the sphere and is the acceleration of the gravity, and 𝛒𝟎 is the density of the liquid. Figure 2: A falling ball viscometer. When the terminal speed is achieved the net force on the sphere is zero, the three forces balance each other out. Looking at the free-body diagram (Figure 1.) this means: 14  𝐖 = 𝐅𝐛 + 𝐅𝐝 (𝟔) Rearranging and regrouping the terms from the above equation the following relationship will be arrived the dynamic viscosity of the liquid: 𝟐𝐫 𝟐 (𝛒 − 𝛒𝟎 )𝐠 𝛍= (𝟕) 𝟗𝐯𝐟 From Eq. (7) the terminal speed is obtained as follows: 𝟐𝐫 𝟐 (𝛒 − 𝛒𝟎 )𝐠 𝐯𝐟 = (𝟖) 𝟗𝛍 The equation above gives a straight line based on the formula 𝐲 = 𝐦𝐱, where the variable 𝑥 offset 𝐫 𝟐 and the variable offset 𝐯𝐟. The slope of this line is: 𝟐(𝛒 − 𝛒𝟎 )𝐠 𝐬𝐥𝐨𝐩𝐞 = (𝟗) 𝟗𝛍 Finally, the dynamic viscosity (𝛍) of the liquid as a function of the slope reads: 𝟐(𝛒 − 𝛒𝟎 )𝐠 𝛍= (𝟏𝟎) 𝟗 × 𝐬𝐥𝐨𝐩𝐞 It should be noted that in actual practice the experiment is performed in a liquid column of finite depth, thus the terminal speed 𝐯𝐟 can be computed using: 𝐋 𝐯𝐟 = (𝟏𝟏) 𝐭 where 𝐭 is the time taken by the ball to drop a distance 𝐋 (distance between two marker) with the recording started after it has reached the terminal speed. Data: In our case, the cylinder is filled with gylcerin with density 𝛒𝟎 = 𝟏𝟐𝟔𝟎 𝐤𝐠/𝐦𝟑 , the density of the sphere ball is 𝛒 = 𝟕𝟕𝟖𝟎 𝐤𝐠/𝐦𝟑 , and the distance between the two markers 𝐋 = 𝟔𝟎 𝐜𝐦. 15  𝐫 (× 𝟏𝟎−𝟑 𝐦) 𝐫 𝟐 (× 𝟏𝟎−𝟔 𝐦𝟐 ) 𝐭 (𝐬) 𝐯𝐟 (× 𝟏𝟎−𝟐 𝐦/𝐬) 1 ………………… 19.78 ………………… 1.5 ………………… 14.89 ………………… 2 ………………… 11.11 ………………… 3 ………………… 6.38 ………………… i. Complete the table above ii. Plot a graph of 𝐯𝐟 (y-axis) against 𝐫 𝟐 (x-axis). Use the graph paper on page 17 (Display Titles and units of x and y-axis) iii. Determine the slope of the graph 𝒚𝟐 − 𝒚𝟏 … − … 𝐬𝐥𝐨𝐩𝐞 = = =⋯ 𝐦−𝟏 𝐬 −𝟏 𝒙𝟐 − 𝒙𝟏 … − … iv. Calculate the experimental value of the dynamic viscosity of glycerin, 𝝁𝐞𝐱𝐩 and write its unit 𝟐(𝛒 − 𝛒𝐨 )𝐠 𝟐(𝟕𝟕𝟖𝟎 − 𝟏𝟐𝟔𝟎) × 𝟗. 𝟖 𝛍𝐞𝐱𝐩 = = =⋯ 𝐏𝐚. 𝐬 𝟗 × 𝐬𝐥𝐨𝐩𝐞 𝟗×… v. Calculate the error percent between theoretical (𝛍𝐭𝐡 = 𝟏. 𝟕 𝐤𝐠/𝐦. 𝐬) and experimental (𝜇exp ) values of dynamic viscosity of a given liquid glycerin |𝛍𝐞𝐱𝐩 − 𝛍𝐭𝐡 | | … − 𝟏. 𝟕| 𝐞𝐫𝐫𝐨𝐫 % = × 𝟏𝟎𝟎 = × 𝟏𝟎𝟎 = ⋯ % 𝛍𝐭𝐡 𝟏. 𝟕 16  (Please use the graph paper below) Check: 1. In the viscosity experiment, which method is used to calculate the dynamic viscosity of the liquid? a) momentum balance b) energy analysis c) force balance d) conservation of mass 2. Th SI unit of the dynamic viscosity is b) Kg. m. s −1 b) Kg. m−1. s −2 c) Kg. m2. s −1 d) Kg. m−1. s −1 3. The frictional effect between different layers of a moving fluid is called a) density b) flow rate c) viscosity d) fluidity 4. When the temperature increases, the viscosity of liquids a) increases b) decreases c) remains constant d) may increase or decrease 5. A steel sphere falling in a liquid experience a retarding force called a) tension force b) Stoke’s force c) drag force d) gravitational force 6. According to Stoke’s law, drag force is given by: a) 𝐹𝑑 = 6πμrv b) 𝐹𝑑 = 6πμrv 2 c) 𝐹𝑑 = 6πμr 2 v d) 𝐹𝑑 = 6π(μrv)2 17 7. When a sphere of radius 𝑟 and density 𝜌 falls through a liquid of density 𝜌0 , it experiences a buoyant force: a) 𝐹𝑏 = 4⁄3 𝜋𝑟 3 𝜌0 𝑔 b) 𝐹𝑏 = 4⁄3 𝜋𝑟 3 𝜌 𝑔 c) 𝐹𝑏 = 4𝜋𝑟 3 𝜌 𝑔 d) 𝐹𝑏 = 6π𝜌0 𝑔 8. If 𝐹𝑑 is the drag force, 𝐹𝑏 is the buoyant force, it is found that when the sphere falling in viscous liquid attains a terminal speed is reached, then it’s weight: a) W = 𝐹𝑏 − 𝐹𝑑 b) W = 𝐹𝑑 − 𝐹𝑏 c) W = 𝐹𝑏 + 𝐹𝑑 d) W = 𝐹𝑏 = 𝐹𝑑 9. The device used for measuring the viscosity of a liquid is a) thermometer b) nanometer c) viscometer d) barometer 10.The terminal velocity of a steel sphere falling in a liquid is 0.03 m/s. The sphere is of radius 𝑟 = 10−3 m and density 𝜌 = 7830 Kg/m3. The density of the liquid is 800 Kg/m3 , then the dynamic viscosity of this liquid a) μ = 1.02 kg⁄m. s b) μ = 2.04 kg⁄m. s c) μ = 0.51 kg⁄m. s d) μ = 10.2 kg⁄m. s 18