[2081 GIE ‘A’] A simple pendulum 2m long swings with an amplitude of 0.1m. Calculate the velocity of the pendulum at its lowest point and its acceleration at extreme ends. [3] Ans: 0.22 m/s and 0.5 m/s2 Solution: Length of pendulum (l) = 2m Amplitude (r) = 0.1m i. Velocity at lowest point (v) =? (Mean position) ii. Acceleration at extreme point (a) =? We know that, [latex]T = 2\pi \sqrt{\frac{l}{g}}[/latex] = [latex]2\pi \sqrt{\frac{2}{10}} = 2.81s[/latex] Again, [latex]\omega = \frac{2\pi}{T} = \frac{2\pi}{2.81}[/latex] = 2.236 rad/s Now, at mean position, [latex]v = \omega r = 2.236\times 0.1 = 0.224[/latex] m/s Again, at extreme position, [latex]a = \omega^2 r = (2.236)^2 \times 0.1 = 0.499[/latex] m/s2
[2081 GIE ‘B’] On an average a human heart is focussed to beat 85 times in a minute. Calculate its frequency and time period. [2] Ans: 1.42 Hz, 0.7 sec Solution: Beats of heart = 85 times per minute Frequency (f) =? Time period (T) =? We know that, f = beats per second = [latex]\frac{85}{60}[/latex] = 1.416 Hz Again, Time period (T) = [latex]\frac{1}{f} = \frac{1}{1.416}[/latex] = 0.706 s
[2081 B/C] One end of a light spring having spring constant 18 N/m is attached to a rigid support. A mass of 0.15 kg is suspended from other end of the spring. A student pulls down it such that extension of spring increases by 4 cm. The student now releases it, as a result, the mass performs SHM. i. Define simple harmonic motion. [1] ii. Calculate the maximum acceleration and time period of SHM. [2] Ans: 48 m/sec2, 0.57 sec Solution: Spring constant (k) = 18 N/m Mass (m) = 0.15 kg Amplitude (r) = 4 cm = 0.04m i. The to and fro motion of an object about their mean position is called simple harmonic motion, where acceleration produced is directly proportional to displacement and is always directed opposite to it. ii. maximum acceleration (a) =? Time period of SHM (T) =? We know, Restoring force (F) = k.r Or, ma = k.r Or, [latex]a = \frac{kr}{m} = \frac{18\times 0.04}{0.15}[/latex] = 48 m/s2 Again. T = [latex]2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{0.15}{18}}[/latex] = 0.573 s
[2081 ‘D’] A simple pendulum of effective length 4 m swings with an amplitude of 0.2 m. Compute the velocity of pendulum at its lowest point. [g = 9.8] [2] Ans: 0.32 m/sec Solution: Length of pendulum (l) = 4m Amplitude (r) = 0.2m Velocity at lowest point (v) =? (Mean position) We know that, [latex]T = 2\pi \sqrt{\frac{l}{g}}[/latex] = [latex]2\pi \sqrt{\frac{4}{10}} = 3.973s[/latex] Again, [latex]\omega = \frac{2\pi}{T} = \frac{2\pi}{3.973}[/latex] = 1.581 rad/s Now, at mean position, [latex]v = \omega r = 1.581\times 0.2 = 0.32[/latex] m/s
[2080 ‘P’] A bob of mass 8 kg performs SHM of amplitude 30 cm. The restoring force is 60 N. Calculate: i. Time period ii. The maximum acceleration iii. Kinetic energy when displacement is 12 cm [3] Ans: (i) 1.256 sec (ii) 7.5 m/sec2 (iii) 7.56 J Solution: Mass (m) = 8kg Amplitude (r) = 30cm = 0.3 m Restoring force (F) = 60N Time period (T) =? Maximum acceleration (a) =? Kinetic energy (KE) =? When y = 12 cm (i) We know, Or, F = k.r Or, k = [latex]\frac{F}{r} = \frac{60}{0.3}[/latex] = 200 N/m Again, T = [latex]2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{8}{200}}[/latex] = 1.256 s (ii) First, [latex]\omega = \frac{2\pi}{T} = \frac{2\pi}{1.256}[/latex] = 5 rad/s Now, acceleration (a) = [latex]\omega^2 r = 5^2 \times 0.3[/latex] = 7.5 m/s2 (iii) When y = 12 cm = 0.12 m Velocity (v) = [latex]\omega \sqrt{r^2 – y^2} = 5\times \sqrt{0.3^2-0.12^2}[/latex] = 1.374 m/s Again, K.E. = [latex]\frac{1}{2}mv^2 = \frac{1}{2}\times 8 \times 1.374^2[/latex] = 7.55 J
[2079 GIE ‘A’] A particle of mass 0.25 Kg oscillates with a period of 2 sec. If its greatest displacement is 0.4 m, what is its maximum kinetic energy? [2] Ans: 0.2 J Solution: Mass (m) = 0.25 kg Time period (T) = 2 s Amplitude (r) = 0.4 m Maximum K.E. =? First, [latex]\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = 3.14 rad/s[/latex] Again, K.E. = [latex]\frac{1}{2}mv^2 = \frac{1}{2}m(\omega r)^2[/latex] = [latex]\frac{1}{2}\times 0.25 \times (3.14\times 0.4)^2[/latex] = 0.197 J
[2079 GIE ‘B’] Calculate the period of oscillation of a simple pendulum of length 1.8 m with a bob of mass 2.2 kg. If the bob of this pendulum is pulled aside a horizontal distance of 20 cm and released, what will be the K.E. of the bob at the lowest point of swing. [3] Ans: 0.245 J Solution: Time period (T) =? Length (l) = 1.8 m Mass (m) = 2.2 kg Amplitude (r) = 20 cm = 0.2 m K.E. =? at Mean Position We know that, T = [latex]2\pi \sqrt{\frac{l}{g}}[/latex] = [latex]2\pi \sqrt{\frac{1.8}{9.8}}[/latex] = 2.693 s Again, [latex]\omega = \frac{2\pi}{T} = \frac{2\pi}{T}[/latex] = [latex]\frac{2\pi}{2.693}[/latex] = 2.3 rad/s Now, K.E. = [latex]\frac{1}{2}mv^2 = \frac{1}{2}\times m(\omega r)^2[/latex] = [latex]\frac{1}{2}\times 2.2 \times (2.3\times 0.2)^2[/latex] = 0.233 J
[2076 ‘GIE’ ‘A’] A particle of mass 0.3 kg vibrates with a period of 2 seconds. If its amplitude is 0.5 m, what is its maximum kinetic energy? Ans: 0.37 J Solution: Mass of particle (m) = 0.3 kg Time period (T) = 2s Amplitude (r) = 0.5 m Maximum K.E. (KEmax) =? First, angular velocity ([latex]\omega[/latex]) = [latex]\frac{2\pi}{T} = \frac{2\times 3.14}{2} = 3.14\ rad.s^{-1}[/latex] Now, velocity is maximum at mean position, So, vmax = [latex]\omega r = 3.14\times 0.5[/latex] = 1.57 m/s Now, K.E.max = [latex]\frac{1}{2}mv^2 = \frac{1}{2}mv^2 = \frac{1}{2}\times 0.3 \times 1.57^2 = 0.369 J[/latex]
[2076 ‘GIE’ ‘B’/ 2074 ‘S’] A simple pendulum 4 m long swings with an amplitude 0.2 m. Compute i. velocity of the pendulum at its lowest point, ii. its acceleration at the end of its path. Ans: 0.32 m/sec, 0.5 m/s2 Solution: Length of pendulum (l) = 4m Amplitude (r) = 0.2 m i.Velocity at lowest point (v) =? (Mean position) ii. Acceleration at extreme point (a) =? We know that, [latex]T = 2\pi \sqrt{\frac{l}{g}}[/latex] = [latex]2\pi \sqrt{\frac{4}{10}} = 3.97 s[/latex] Again, [latex]\omega = \frac{2\pi}{T} = \frac{2\pi}{T}[/latex] = 1.581 rad/s Now, at mean position, [latex]v = \omega r = 1.581\times 0.2 = 0.316[/latex] m/s Again, at extreme position, [latex]a = -\omega^2 r = -(1.581)^2 \times 0.2 = 0.499[/latex] m/s2
[2076 ‘C’]The velocity of a particle executing simple harmonic motion is 16 cms-1 at a distance of 8 cm from the mean position and 8 cms-1 at a distance of 12 cm from the mean position. Calculate the amplitude of the motion. Ans: 0.1308 m Solution: Case – 1: Displacement (y) = 8 cm = 0.08 m Velocity [latex](v_8) = 16 cms^{-1} = 0.16 ms^{-1}[/latex] Case – 2: Displacement (y) = 12 cm = 0.12 m Velocity [latex](v_{12}) = 8 cms^{-1} = 0.08 ms^{-1}[/latex] Amplitude (r) =? We know, Velocity (v) = [latex]\omega \sqrt{r^2 – y^2}[/latex] At y = 8 cm = 0.08 m [latex]v_8 = \omega \sqrt{r^2 – 0.08^2}[/latex] Or, 0.16 = [latex]\omega \sqrt{r^2 – 0.08^2}[/latex] ……………….. (i) Again, at y = 12 cm = 0.12 m [latex]v_{12} = \omega \sqrt{r^2 – 0.12^2}[/latex] Or, 0.08 = [latex]\omega \sqrt{r^2 – 0.12^2}[/latex] …………… (ii) Dividing [latex]eq^{n.}\ (ii)\ by\ eq^{n.}[/latex] (i), we get, [latex]\frac{0.08}{0.16} = \frac{\omega \sqrt{r^2 – 0.12^2}}{\omega \sqrt{r^2 – 0.08^2}}[/latex] Or, [latex]\frac{1}{2} = \frac{\sqrt{r^2 – 0.0144}}{\sqrt{r^2 – 0.0064}}[/latex] Or, [latex]\sqrt{r^2 – 0.0064} = 2\sqrt{r^2 – 0.0144}[/latex] Squaring both sides, we get, [latex]r^2 – 0.0064 = 2(r^2 – 0.0144)[/latex] Or, [latex]r^2 – 0.0064 = 2r^2 – 0.0288[/latex] Or, [latex]2r^2 – r^2 = -0.0064 + 0.0288[/latex] Or, [latex]r^2 = 0.0224[/latex] Or, r = 0.149 m
[2075 ‘B’] The position of a certain object in S.H.M. is given as x = 0.05 cos(290t + 2.5); where x is in meter and t is in sec. What are the amplitude, period and initial phase angle for this motion? Ans: 0.05 m, 0.022 sec, 2.5 rad. Solution: Equation of S.H.M. (x) = 0.05cos(290t + 2.5) Amplitude (r) =? Time period (T) =? Initial Phase angle [latex](\phi)[/latex] =? Comparing above equation with: x = [latex]rcos(\omega t + \phi)[/latex], we get, Amplitude (r) = 0.05 m [latex]\omega[/latex] = 290 rad/s So, [latex]\omega = \frac{2\pi}{T}[/latex] Or, [latex]290 = \frac{2\pi}{T}[/latex] Or, [latex]T = \frac{2\pi}{290}[/latex] = 0.022 s Also, initial phase angle [latex](\phi)[/latex] = 2.5 rad
[2074 ‘A’] A body of mass 0.1 kg is undergoing simple harmonic motion of amplitude 1 m and period 0.2 second. If the oscillation is produced by a spring, what will be maximum value of the force and the force constant of the spring? Ans: 98.6 N, 98.6 N/m Solution: Mass (m) = 0.1 kg Amplitude (r) = 1 m Time period (T) = 0.2 s Force (F) =? Force constant (k) =? We know that, T = [latex]2\pi \sqrt{\frac{m}{k}}[/latex] Or, 0.2 = [latex]2\pi \sqrt{\frac{0.1}{k}}[/latex] Or, [latex]\frac{0.2}{2\pi} = \sqrt{\frac{0.1}{k}}[/latex] Or, 0.032 = [latex]\sqrt{\frac{0.1}{k}}[/latex] Squaring both sides, [latex](1.013\times 10^{-3}) = \frac{0.1}{k}[/latex] Or, k = [latex]\frac{0.1}{1.013\times 10^{-3}}[/latex] = 98.69 N/m Again, F = k.r Or, F = [latex]98.69 \times 1[/latex] = 98.69 N
[2072 ‘E’] A body of mass 2 kg is suspended from a spring of negligible mass and is found to stretch the spring 0.1 m. What is its force constant and the time period? Ans: 200 N/m, 0.63 sec Solution: Mass (m) = 2 kg Amplitude (r) = 0.1 m Force constant (k) =? Time period (T) =? We know that, Force (F) [latex]\propto r[/latex] Or, F = k.r Or, mg = k.r Or, k = [latex]\frac{mg}{r} = \frac{2\times 10}{0.1}[/latex] = 200 N/m Again, Time period (T) = [latex]2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{2}{200}}[/latex] = 0.63 s
