Tiêu đề 8. P1C8 Periodic Motion_With Solve.pdf ✅

2  Physics 1st Paper Chapter-8  MwZi ïiæ we ̄Ívi †_‡K ZvB, x = A cos(t)  x = 1.5  10–4 cos(2  262t) v = dx dt = – 1.5  10–4  2  262 sin(2  262t) v = – 0.0786 sin(524t)  a = dv dt  a = – 0.0786  524 cos(524t)  a = –41.1864 2 cos(524t) (Ans.) 7| hLb 1 kg Av`k© fi GKwU Pjgvb cøvUd‡g©i Dci ivLv nq ZLb Zvi ̄ú›`‡bi nvi 125 vib min–1 ARvbv KZ f‡ii Rb ̈ ̄ú›`‡bi nvi 243 vib min–1 n‡e? Pjgvb cøvUd‡g©i fi AMÖvn ̈ Ki| [BUET 10-11] mgvavb: T = 2 m k = 1 f  f  1 m  f1 f2 = m2 m1 = 125 243  125 243 = m2 1  m2 = 0.2646 kg (Ans.) 8| †m‡KÛ †`vj‡Ki •`N© ̈ 1% e„w× Ki‡j, D3 †`vjK w`‡b KZ mgq nviv‡e? [BUET 07-08] mgvavb: T  L [1% e„w× Kiv A_© Avw` •`N© ̈ 100 GKK n‡j bZzb •`N© ̈ 101 GKK]  T T = 86400 86400 – x = 101 100  x = 428.78677 s (Ans.) 9| GKwU †m‡KÛ †cÛzjvgwewkó Nwo cÖwZw`b Avav wgwbU (30 sec) jvf K‡i| †cÛzjvgwU mwVKmgq w`‡Z n‡j Dnvi mij †`vj‡Ki •`‡N© ̈i Kx cwigvY n«vm-e„w× NUv‡Z n‡e? (g = 980 cm–2 ) [BUET 07-08] mgvavb: bZzb •`N© ̈ L2 n‡j, L2 L1 =     T2 T1 2 =       2 2  86400 86400 + 30 2 =     86400 + 30 86400 2  L2 – L1 L1  100% = 0.0694%  •`N© ̈ e„w× Ki‡Z n‡e 0.0694%| (Ans.) 10| mijQw›`Z ̄ú›`bm¤úbœ GKwU e ̄‘i †eM 3 ms–1 hLb miY 4 m Ges †eM 4ms–1 hLb miY 3m| (a) †`vj‡bi we ̄Ívi I ch©vqKvj wbY©q Ki| (b) e ̄‘wUi fi 50 kg n‡j †`vj‡bi †gvU kw3 wbY©q Ki| [BUET 05-06] mgvavb: (a) v1 =  A 2 – x 2 1  3 =  A 2 – 16 .......... (i) v2 =  A 2 – x 2 2  4 =  A 2 – 9 .......... (ii) 3 4 = A 2 – 16 A 2 – 9  9 16 = A 3 – 16 A 2 – 9 [Calculator-G A 2 †K x a‡i]  A 2 = 25  A = 5 m (Ans)  3 =  25 – 16   = 1  T = 2  = 2 s (Ans.) (b) Etotal = 1 2 m 2 A 2 = 1 2  50  1  5 2 = 625 J (Ans) 11| GKwU IRb gvcvi w ̄úas wbw3i Dci `vov‡bvi ci Zzwg jÿ ̈ Ki‡j †h mvg ̈v ̄’vq Avmvi c~‡e© wbw3i KuvUvwU mvg ̈ve ̄’vi `ycv‡k K‡qKevi †`vj Lvq| †`vjbKvj 0.8 †m‡KÛ n‡j Ges †Zvgvi fi 64 kg n‡j wbw3i w ̄úas aaæeK KZ? [BUET 01-02] mgvavb:  = 2 T = 2 0.8 = 5 2  rads–1 k = m 2 = 64      5 2 2 = 3947.84176 Nm–1 (Ans.) 12| GKwU e ̄‘i Qw›`Z MwZ x = 6.0 cos(3t + /3) m mgxKiY Øviv wee„Z Kiv hvq| t = 2 sec mg‡q (i) miY (ii) †eM Ges (iii) Z¡iY †ei K‡iv| [BUET 00-01] mgvavb: (i) x = 6 cos    3  2 +  3 m = 3 m (ii) v = dx dt = – 6  3 sin    6 +  3 = –48.98 ms–1 (iii) a = dv dt = – 6  3  3 cos    3  2 +  3 = –266.548 ms–2 (Ans.) 13| AwfKl©R Z¡iY g = 9.8 ms–2 Gi Rb ̈ GKwU †m‡KÛ †`vj‡Ki •`N© ̈ wbY©q K‡iv| AwfKl©R Z¡iY 10% Kwg‡q D3 †`vj‡Ki †`vjbKvj wVK ivL‡Z n‡j •`‡N© ̈i Kx cwieZ©b n‡e? [KUET 19-20] mgvavb: T = 2 L g  L = gT2 4 2