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OSCILLATIONS 14.19, , , , , , , , (As G=E,, (At =wyed ky? or ay? =A?, A, or yet =+071A, v2, , 0.71 times the amplitude on either side of, mean position., 1, (i) Here, 9 =, , 5 “max, , In general, kinetic energy, , , , , , , , , , 1 al ) Pi) 2, =< m| — 2, = mv,, 2 2 hg Pima) gg Vmax, 1 ', = x Maximum kinetic energy, 1, or ral (2), From equation (1),, =} k (A -y?), Vaz, (Eda = 5 FA [Put y =0], , Putting these values in equation (2), we get, eae oe 1. IR >, ae ed, 2 (A°-y*) 7% ace, , or Wy? =3A, 3, 3 4=4086A, 2, =0.86 times the amplitude on either side of, mean position, , , , +, , , , % PROBLEMS FOR PRacTICE, , L_A bob of simple pendulum of mass 1 g is oscillating, with a frequency 5 vibrations per second and its, amplitude is 3 cm. Find the kinetic energy of the, bob in the lowest position. (Ans. 4441.5 erg), , 2 A body weighing 10 g has a velocity of 6cms”!, ae one second of its starting from mean position., , ee time Period is 6 seconds, find the kinetic, "BY, potential energy and the total energy., , (Ans. 180 erg, 540 erg, 720, en . y b erg), sr beaeas executes SHM of Period 8 seconds. After, ee ae is peeing through the mean position, , BY be half kinetic and half potential ?, [Chandigarh 08}, (Ans. 1's), , 2, , Samsung Triple Camera, Shot with my Galaxy M21, , A. The total energy of a particle executing SHM of, period 2n seconds is 1.024 * 10 "J, The displacement of the partic le at t/ 4s is 0.08/2 m Calculate, , f motion and mass of the particle., , (Ans. 0.16 m ; 0.08 kg), , , , the amplitude, , ached to a spring oscillates, armonic motion with a, | energy of 10 J. If the, maximum speed of the pi | what is, the force constant of the spring ? What will be the, ential energy of the spring during the, , (Ans. k= 500 Nm!, Uppy, = 105), , 5. A particle which is att, horizontally with simple hi, , frequency of 1/ Hz and total, article is 0.4 ms, , maximum pot, motion ?, , 6. The length of a weightless spring increases by 2 cm, , when a weight of 1.0 kg is suspended from it. The, weight is pulled down by 10 cm and released, Determine the period of oscillation of the spring, , and its kinetic energy of oscillation, si (Ans. 0.28 s,25J), , , , % HINTS, , 1. At the lowest or the mean position, energy of the, bob is entirely kinetic and maximum., , (Ed pan = ; mo? A?, 1 ap ta ashi ce, = 5 m(2nvy A? = 2x mv" A, , = 2987 x1x5? x3? = 44415 erg., , , , 2. Herem=10g, T, , , , Ga22, Te) va, When f=1s, v=6cms7, , As v=Aacos ot, , 6=Ax cose x1=Ax= ., ies 1=A qe, , , , =Ax or A= cn, x, Total energy,, 1 36, ==x10x|— =, ig ( = \ = 720 erg., , , , , , 1 2, Kinetic energy == imu? = — x 10 x6" = 180 erg,, , 1, , Potential energy, , = Total energy ~ Kinetic energy, , , , , , ) ~ 180 = 540 erg., 3. As =, vw), or en, 2