Tiêu đề 9. P1C9 Mastery Practice Sheet তরঙ্গ (With Solve) .pdf ✅

4  Physics 1st Paper Chapter-9 Rhombus Publications  f2 – f1 = 3 [∵ f2 > f1]  1.01 f1 – f1 = 3  .01 f1 = 3  f1 = 300 Hz Ges f2 = 303 Hz (Ans.) 21. MKS unit G GKwU AMÖMvgx Zi‡1⁄2i mgxKiY y = 0.5 sin    200t –  20x 13 wbY©q Ki, (i) K¤úv1⁄4 (ii) Zi1⁄2 •`N© ̈ Ges (iii) Zi1⁄2 †eM| [BUET 04-05] mgvavb: y = 0.5 sin    200t –  20x 13 = 0.5 sin      20  13 (130t – x) = 0.5 sin       2 13 10 (130t – x) ......(i) Avgiv Rvwb, y = a sin 2  (vt – x) ......(ii) (i) I (ii) Zzjbv K‡i cvB,  = 13 10 = 1.3 m; v = 130 ms–1 GLv‡b, (i) K¤úv1⁄4; f = v  = 130 1.3 = 100 Hz (Ans.) (ii) Zi1⁄2 •`N© ̈,  = 1.3 m (Ans.) (iii) Zi1⁄2 †eM, v = 130 ms–1 (Ans.) 22. evqy‡Z I cvwb‡Z 300 Hz K¤úvs‡Ki GKwU kã Zi1⁄2 •`‡N© ̈i cv_©K ̈ 4.16 wgUvi| evqy‡Z k‡ãi †eM 352 ms–1 n‡j cvwb‡Z k‡ãi †eM KZ? [BUTex 05-06, RUET 04-05] mgvavb: cvwb‡Z k‡ãi †eM †ewk, ZvB w > a w – a = 4.16  vw f – va f = 4.16  vw f – va f = 4.16  vw – va = 4.16  f [f = 300 Hz]  vw = 352 + 4.16  300 = 1600 ms–1 (Ans.) 23. ÒAbybv`Ó Gi msÁv `vI| ev ̄Íe Rxe‡b Gi mdj cÖ‡qv‡Mi GKwU D`vniY `vI| [KUET 04-05] mgvavb: †Kvb e ̄‘i wbR ̄^ K¤úv1⁄4 Zvi Dci Av‡ivwcZ ch©ve„Ë ̄ú›`‡bi K¤úv‡1⁄4i mgvb n‡j e ̄‘wU m‡e©v”P we ̄Ív‡i Kw¤úZ nq, Zv‡K Abybv` e‡j| evqy‡Z k‡ãi MwZ‡eM wbY©‡q Abybv‡`i aviYv e ̈eüZ nq| 24. †Kvb GKwU gva ̈‡g GKwU myi kjvKv n‡Z Drcbœ k‡ãi Zi1⁄2 •`N© ̈ 3 cm Ges GKB gva ̈‡g k‡ãi †eM 330 ms–1 n‡j H gva ̈‡g myi kjvKvwUi 55wU c~Y© K¤ú‡b kã KZ `~i hv‡e? [BUTex 04-05] mgvavb: S = N = 55  3 100 = 1.65 m (Ans.) 25. GKwU AMÖMvgx Zi‡1⁄2i mgxKiY n‡”Q y = 100 sin     x  100 – t 0.25 | GLv‡b me KqwU ivwk SI GK‡K n‡j Zi1⁄2wUi Zi1⁄2 •`N© ̈, ch©vqKvj, K¤úvsK Ges †eM wbY©q Ki| [KUET 04-05] mgvavb: y = 100      x 100 – t 0.25  y = – 100 sin      t 0.25 – x 100  y = – 100 sin  100     t  100 0.25 – x  y = – 100 sin  100 (400t – x) ......... (i) Avgiv Rvwb, y = A sin 2  (vt – x) ......... (ii) (i) I (ii) Zzjbv K‡i cvB,  2  =  100   = 200 m (Ans.) 400t = vt  v = 400 m/s (Ans.)  v = f  f = v   f = 400 200  f = 2 Hz (Ans.)  T = 1 f  T = 1 2 = 0.5 s (Ans.) 26. A I B †K GKB mv‡_ Kuvc‡j cÖwZ †m‡K‡Û 5wU ̄^iK‡¤úi m„wó nq| hw` A Gi Dci wKQzUv IRb †`qv hvq, Z‡e ̄^iK‡¤úi msL ̈v K‡g hvq| hw` B Gi K¤úv1⁄4 cÖwZ †m‡K‡Û 256 nq, Z‡e A Gi K¤úv1⁄4 wbY©q Ki| [BUET 03-04] mgvavb: fA  fB = 5 ....... (i) A †Z IRb w`‡j ̄^iK‡¤úi msL ̈v K‡g hvq ZvB, fA > fB  fA – fB = 5 †`Iqv Av‡Q, B-Gi K¤úv1⁄4, fB = 25 Hz awi, A-Gi K¤úv1⁄4 = fA  fA = 5 + fB = 5 + 256  fA = 261 Hz (Ans.)