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Getaran sebuah kawat ditentukan oleh persamaan: Tunjukkan bahwa persamaan tersebut ekuivalen dengan persamaan: y = 4 A cos T 2 π t ⋅ cos λ 2 π x

Getaran sebuah kawat ditentukan oleh persamaan:

y equals 2 A open parentheses cos open square brackets 2 straight pi begin italic style left parenthesis straight t over straight T minus straight x over straight lambda right parenthesis end style close square brackets plus cos open square brackets 2 straight pi open parentheses t over T italic plus x over lambda close parentheses close square brackets close parentheses

Tunjukkan bahwa persamaan tersebut ekuivalen dengan persamaan:

$y = 4 A cos \frac{2 π t}{T} \cdot cos \frac{2 π x}{λ}$

N. Puspita

Master Teacher

Jawaban terverifikasi

Pembahasan

Dengan menggunakan rumus trigonometri, diperoleh: Dengan demikian, terbukti bahwa

Dengan menggunakan rumus trigonometri, diperoleh:

$table attributes columnalign right center left columnspacing 0px end attributes row y equals cell 2 A open parentheses cos open square brackets 2 straight pi begin italic style left parenthesis t over T minus x over lambda right parenthesis end style close square brackets plus cos open square brackets 2 straight pi open parentheses t over T italic plus x over lambda close parentheses close square brackets close parentheses end cell row blank equals cell 2 A open parentheses 2 cos space 1 half open square brackets 2 straight pi open parentheses t over T italic minus x over lambda close parentheses plus 2 straight pi open parentheses straight t over straight T italic plus straight x over straight lambda close parentheses close square brackets times cos space 1 half open square brackets 2 straight pi open parentheses t over T italic minus x over lambda close parentheses minus 2 straight pi open parentheses straight t over straight T italic plus straight x over straight lambda close parentheses close square brackets close parentheses end cell row blank equals cell 2 A open parentheses 2 cos space 1 half open square brackets fraction numerator 2 straight pi t over denominator T end fraction minus fraction numerator 2 pi x over denominator straight lambda end fraction plus fraction numerator 2 pi t over denominator T end fraction plus fraction numerator 2 pi x over denominator straight lambda end fraction close square brackets times cos space 1 half open square brackets fraction numerator 2 straight pi t over denominator T end fraction minus fraction numerator 2 pi x over denominator straight lambda end fraction minus fraction numerator 2 pi t over denominator T end fraction minus fraction numerator 2 pi x over denominator straight lambda end fraction close square brackets close parentheses end cell row blank equals cell 2 A open parentheses 2 cos space open parentheses 1 half times fraction numerator 4 straight pi t over denominator T end fraction close parentheses times cos space open parentheses 1 half times fraction numerator negative 4 πx over denominator straight lambda end fraction close parentheses close parentheses end cell row blank equals cell 2 A open parentheses 2 cos fraction numerator 2 straight pi t over denominator T end fraction times cos fraction numerator 2 straight pi x over denominator straight lambda end fraction close parentheses end cell row blank equals cell 4 A cos fraction numerator 2 straight pi t over denominator T end fraction times cos fraction numerator 2 straight pi x over denominator straight lambda end fraction end cell row blank blank blank end table$

Dengan demikian, terbukti bahwa

$2 A open parentheses cos open square brackets 2 straight pi open parentheses straight t over straight T italic minus straight x over straight lambda close parentheses close square brackets plus cos open square brackets 2 straight pi open parentheses t over T italic plus x over lambda close parentheses close square brackets close parentheses equals 4 A cos space fraction numerator 2 pi t over denominator T end fraction times cos fraction numerator 2 pi x over denominator lambda end fraction$