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**Biju Patnaik University of Technology BPUT - BPUT****1282 Views**- 57 Offline Downloads
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B. TECH PHYSICS LABORATORY MANUAL GOVERNMENT COLLEGE OF ENGINEERING, KALAHANDI, BHAWANIPATNA Mrs. A. Gupta Dr. S.K. Mohanta Department of Basic Science ©Dept. of Basic Science, GCEK, Bhawanipatna

Contents 1. Acceleration due to gravity (g) by bar (Compound) pendulum. 2. Determination of wavelength of Sodium Yellow light using Newton’s ring set-up. 3. Young’s modulus by Searl’s apparatus. 4. Determination of Grating element. 5. Hall effect. 6. Surface tension by Capillary rise method. 7. Young’s modulus by bending of beam. 8. Rigidity modulus by Static method. 9. P-N diode charactreistics. 10. Characteristics of BJT.

EXPERIMEN NO. 1 Aim: To determine the acceleration due to gravity (g) by using a Bar pendulum and also determine radius of gyration about an axis through the center of gravity for the bar pendulum. Apparatus Required: (i) A bar pendulum, (ii) a knife–edge with a platform, (iii) a sprit level, (iv) a precision stop watch, and (v) a meter scale. Theory: A simple pendulum consists of a small body called a “bob” (usually a sphere) attached to the end of a string the length of which is great compared with the dimensions of the bob and the mass of which is negligible in comparison with that of the bob. Under these conditions the mass of the bob may be regarded as concentrated at its center of gravity, and the length of the pendulum is the distance of this point from the axis of suspension. When the dimensions of the suspended body are not negligible in comparison with the distance from the axis of suspension to the center of gravity, the pendulum is called a compound, or physical, pendulum. A rigid body mounted upon a horizontal axis so as to vibrate under the force of gravity is a compound pendulum. In Fig.1 a body of irregular shape is pivoted about a horizontal frictionless axis through P and is displaced from its equilibrium position by an angle 𝜃. In the equilibrium position the center of gravity G of the body is vertically below P. The distance GP is l and the mass of the body is m. The restoring torque for an angular displacement 𝜃 is 1

For small amplitudes (𝜃 ≈ 0), 𝜏 = − 𝑔 sin 𝜃 𝐼 2𝜃 2 …………….. = − 𝑔𝜃 …………….. where I is the moment of inertia of the body through the axis P. Eq. (2) represents a simple harmonic motion and hence the time period of oscillation is given by 𝐼 𝑔 𝑇 = 𝜋√ .….………… Now 𝐼 = 𝐼𝐺 + , where 𝐼𝐺 is the moment of inertia of the body about an axis parallel with axis of oscillation and passing through the center of gravity G. ……………..(4) 𝐼𝐺 = where K is the radius of gyration about the axis passing through G. Thus, 𝑇 = 𝜋√ 2 2 + 𝑔 𝐾2 + 𝑙 𝜋√ = ……………. 𝑔 The time period of a simple pendulum of length L, is given by 𝑇 = 𝜋√ Comparing with Eq. (5) we get = + ……………(6) 𝑔 2 …………….. This is the length of “equivalent simple pendulum”. If all the mass of the body were concentrated 2 at a point O (See Fig.1) such that 𝑃 = + , we would have a simple pendulum with the same time period. The point O is called the ‘Centre of Oscillation’. Now from Eq. (7) − + = ……………… i.e. a quadratic equation in . Equation (8) has two roots And = + = and such that ……………(9) Thus both 𝑎 are positive. This means that on one side of C.G there are two positions of the centre of suspension about which the time periods are the same. Similarly, there will be a pair of positions of the centre of suspension on the other side of the C.G about which the time periods will be the same. Thus there are four positions of the centers of suspension, two on either side of the C.G, about which the time periods of the pendulum would be the same. The distance between two such positions of the centers of suspension, asymmetrically located on either side of C.G, is the length L of the simple equivalent pendulum. Thus, if the body was supported on a parallel axis through the point O (see Fig. 1), it would oscillate with the same time period T as when 2

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