Tiêu đề 8. P1C8 Periodic Motion_With Solve.pdf ✅

ch©ve„Ë MwZ  Mastery Practice Sheet 3 = 9.8  2 2 4 2 = 0.9929 m = 99.29 cm  L  g  g 10% Kgv‡j AwfKl©R Z¡iYI 10% Kg‡e| †`vj‡Ki •`‡N© ̈i cwieZ©b = 99.29  10% = 9.929 cm (Ans.) 14| GKwU †m‡KÛ †`vjK Nwo cvnv‡oi cv`‡`‡k wVK mgq †`q wKš‘ cvnv‡oi P‚ovq DVv‡j 2 NÈvq 8 †m‡KÛ mg‡qi cv_©K ̈ †`Lvq| c„w_exi e ̈vm 12800 km n‡jÑ [KUET 17-18] (i) cvnv‡oi D”PZv wbY©q K‡iv| (ii) cvnv‡oi P‚ovq mwVKfv‡e KvR Ki‡Z n‡j †`vj‡Ki •`N© ̈ KZ % cwieZ©b Ki‡Z n‡e? mgvavb: (i) 2 NÈvq 8 s ax‡i P‡j  24 NÈvq (8  12) s ax‡i P‡j = 96 s T T = 86400 86400 – x Avevi, T  1 g  R  T T = 86400 86400 – x = R + h R  86400 86400 – 96 = 6.4  106 + h 6.4  106  h = 7119.02 m  L = 0.998889 L (ii) T = 2 L g  T  L  L L =    T   T 2 =     86400 86400 – 96 2 =     900 899 2  L – L L =     900 899 2 – 1 = – 2.2259  10–3  L – L L  100% = –0.222593%  •`N© ̈ n«vm Ki‡Z n‡e 0.222593%| 15| GKwU wjd‡Ui Qv` †_‡K GKwU mij †`vjK Szjv‡bv Av‡Q| wjdU Pjvi mgq GB †`vj‡Ki †`vjbKvj wjd‡Ui w ̄’i Ae ̄’vi Zzjbvq hw` A‡a©K nq, Z‡e wjd‡Ui Z¡i‡Yi w`K I gvb wbY©q K‡iv| [KUET 05-06] mgvavb: T  1 g T2 T1 = g1 g2 = 1 2  g1 g2 = 1 4  g2 > g1  wjdU Dc‡ii w`‡K MwZkxj, Z¡iY a n‡j, g g + a = 1 4  a = 29.4 ms–2 (Ans.) 16| GKwU cvnv‡oi cv`‡`‡k GKwU †m‡KÛ †`vjK mwVK mgq †`q| GwU‡K cvnv‡oi m‡e©v”P k„‡1⁄2 wb‡q †M‡j cÖwZw`b 2 wgwbU ax‡i P‡j| cvnv‡oi D”PZv wbY©q K‡iv| (c„w_exi e ̈vmva© 6400 km) [KUET 04-05] mgvavb: T T = 86400 86400 – 2  60 = g g = R + h R  86400 86400 – 120 = 6.4  106 + h 6.4  106    ∵  g g =     R R + h 2  h = 8901.251739 m (Ans.) 17| c„w_exc„‡ô GKwU mij †`vj‡Ki †`vjbKvj 2 sec| G‡K P›`ac„‡ô wb‡j Gi †`vjbKvj nq 4.5 sec| c„w_exi fi I P‡›`ai f‡ii AbycvZ 81 n‡j c„w_exi e ̈vmva© I P‡›`ai e ̈vmv‡a©i AbycvZ wbY©q K‡iv| [RUET 11-12, 03-04] mgvavb: T = 2 L g ; g = GM R 2  g  M R 2  T  1 g  R M  Tearth Tmoon = Rearth Rmoon  Mmoon Mearth  Rearth Rmoon = 2 4.5  81 = 4 (Ans.) 18| hw` †Kvb ̄’v‡b GKwU †m‡KÛ †`vj‡Ki •`N© ̈ 1 m nq, Z‡e †h †`vjK †mB ̄’v‡b wgwb‡U 20 evi †`vj †`q, Zvi •`N© ̈ †ei K‡iv| [RUET 10-11] mgvavb: 20 evi †`vj †`q 60 s G  1 Ó Ó Ó 60 20 s G = 3 s G  T2 = 3, T1 = 2s ∵ T  L  T2 T1 = L2 L1  L2 =     T2 T1 2  L1  L2 = 9 4  L1  L2 = 2.25 (Ans.) 19| GKwU e ̄‘i Qw›`Z MwZ x = 10.0 cos     5t +  4 m mgxKi‡Yi mvnv‡h ̈ cÖKvk Kiv hvq| t = 2s mg‡q D3 e ̄‘i (K) miY (L) †eM I (M) Z¡iY wbY©q K‡iv| [RUET 09-10] mgvavb: x = 10.0 cos    5t +  4 m (K) t = 2s G,