Benda M melakukan gerak harmonik sederhana. Saat simpangannya P, kecepatannya Q. Saat simpangannya R, kecepatannya S, maka ω2 dapat dinyatakan sebagai ....

Pertanyaan

Benda M melakukan gerak harmonik sederhana. Saat simpangannya P, kecepatannya Q. Saat simpangannya R, kecepatannya S, maka omega squared dapat dinyatakan sebagai ....

  1. open square brackets fraction numerator Q minus P over denominator R minus S end fraction close square brackets squared 

  2. open square brackets fraction numerator Q minus P over denominator R minus S end fraction close square brackets 

  3. open square brackets fraction numerator Q squared minus S squared over denominator R squared minus P squared end fraction close square brackets 

  4. open square brackets fraction numerator Q squared minus S squared over denominator P squared minus S squared end fraction close square brackets 

  5. open square brackets fraction numerator Q squared minus S squared over denominator P squared minus R squared end fraction close square brackets 

Y. Maghfirah

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C.

Pembahasan

Jawaban yang benar untuk pertanyaan tersebut adalah C.

Diketahui :

y subscript 1 equals P v subscript 1 equals Q y subscript 2 equals R v subscript 2 equals S

Ditanya : omega squared

Kecepatan partikel yang bergerak harmonis sedehana dapat dihitung menggunakan persamaan 

v equals omega square root of A squared minus y squared end root

Keadaan pertama

table attributes columnalign right center left columnspacing 0px end attributes row cell v subscript 1 end cell equals cell omega square root of A squared minus y subscript 1 squared end root end cell row Q equals cell omega square root of A squared minus P squared end root end cell row cell Q over omega end cell equals cell square root of A squared minus P squared end root end cell row cell Q squared over omega squared end cell equals cell A squared minus P squared end cell row cell A squared end cell equals cell Q squared over omega squared plus P squared space....... left parenthesis 1 right parenthesis end cell end table

Keaddan ke dua

v subscript 2 equals omega square root of A squared minus y subscript 2 squared end root S equals omega square root of A squared minus R squared end root S over omega equals square root of A squared minus R squared end root S squared over omega squared equals A squared minus R squared A squared equals R squared plus S squared over omega squared space......... left parenthesis 2 right parenthesis

Substitusikan persamaan (1) dan (2)

table attributes columnalign right center left columnspacing 0px end attributes row cell A squared end cell equals cell A squared end cell row cell Q squared over omega squared plus P squared end cell equals cell R squared plus S squared over omega squared end cell row cell Q squared over omega squared minus S squared over omega squared end cell equals cell R squared minus P squared end cell row cell fraction numerator Q squared minus S squared over denominator omega squared end fraction end cell equals cell R squared minus P squared end cell row cell omega squared end cell equals cell fraction numerator Q squared minus S squared over denominator R squared minus P squared end fraction end cell end table 

Jadi, jawaban yang tepat adalah C.

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Jawaban terverifikasi

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