Iklan

Iklan

Pertanyaan

Sebuah bandul dengan panjang tali I diayun dengan sudut simpangan awal θ 0 ​ sehingga berosilasi harmonik. Diketahui panjang tali bandul dijadikan dua kali panjang semula dan bandul dipindahkan kesuatu planet seukuran bumi dengan massa 2 kali massa bumi. Jika bandul itu diberi simpangan awal yang akan terjadi adalah .... (1) Frekuensi osilasinya tetap (2) Selisih antara energi kinetik dan energi potensialpada titik tengah antara titik setimbang dan titik simpangan maksimum adalah4 mgl ( 2 sin θ 0 ​ − 1 ) (3) Energi mekaniknya membesar menjadi 4 kali (4) Perioda osilasi bertambah besar

Sebuah bandul dengan panjang tali I diayun dengan sudut simpangan awal  sehingga berosilasi harmonik. Diketahui panjang tali bandul dijadikan dua kali panjang semula dan bandul dipindahkan ke suatu planet seukuran bumi dengan massa 2 kali massa bumi. Jika bandul itu diberi simpangan awal begin mathsize 14px style straight theta subscript 0 end style yang akan terjadi adalah .... 
(1) Frekuensi osilasinya tetap
(2) Selisih antara energi kinetik dan energi potensial pada titik tengah antara titik setimbang dan titik simpangan maksimum adalah 4mgl
(3) Energi mekaniknya membesar menjadi 4 kali 
(4) Perioda osilasi bertambah besar

  1. (1), (2), dan (3) benar

  2. (1) dan (3) benar

  3. (2) dan (4) benar

  4. Hanya (4) yang benar

  5. Semuanya benar

Iklan

A. Aulia

Master Teacher

Jawaban terverifikasi

Iklan

Pembahasan

begin mathsize 14px style left parenthesis 1 right parenthesis space Benar.  space space space space space space Hitung space terlebih space dahulu space percepatan space gravitasi space planet space  space space space space space space tersebut comma space yaitu space colon  space space space space space space space space space straight g subscript straight b over straight g subscript straight p equals straight M subscript straight b over straight M subscript straight p open parentheses straight R subscript straight P over straight R subscript straight b close parentheses squared  space space space space space space space space space straight g over straight g subscript straight p equals fraction numerator straight M subscript straight b over denominator 2 straight M subscript straight b end fraction open parentheses straight R over straight R close parentheses squared  space space space space space space Sehingga space perbandingan space periode space osilasi space di space bumi space dan space  space space space space space space di space planet space tersebut space adalah space colon  space space space space space space space space space straight T subscript straight b over straight T subscript straight p equals fraction numerator 2 straight pi square root of straight L over straight g end root over denominator 2 straight pi square root of fraction numerator 2 straight L over denominator 2 straight g end fraction end root end fraction equals fraction numerator square root of 1 over denominator square root of 1 end fraction  space space space space space space Maka space colon  space space space space space space space space space straight T subscript straight p equals straight space straight T subscript straight b  space space space space space space Sehingga straight space colon  space space space space space space space space space straight f subscript straight p equals straight f subscript straight b    left parenthesis 2 right parenthesis space Salah.  space space space space space space Energi space mekanik space osilasi space di space setiap space titik space dirumuskan space  space space space space space space dengan space persamaan space colon  space space space space space space space space space EM equals 1 half kA squared straight space  space space space space space space space space space EM equals 2 mπ squared space straight f squared space straight A squared  space space space space space space Titik straight space tengah comma straight space artinya straight space colon  space space space space space space space space space straight theta equals 45 to the power of 0 equals straight pi over 4  space space space space space space Ingat space disini space saat space panjang space bandul space saat space dibawa space ke space  space space space space space space planet space lain.  space space space space space space Energi space kinetiknya space colon  space space space space space space space space space EK equals 1 half mv squared space cos squared invisible function application space open parentheses straight theta subscript straight t close parentheses equals mg space 2 space straight l space cos squared invisible function application space open parentheses straight theta subscript 0 close parentheses  space space space space space space space space space EK equals 2 mgl open parentheses 1 minus sin squared invisible function application space straight theta subscript 0 close parentheses  space space space space space space space space space EK equals 2 mgl minus 2 mgl space sin squared space straight theta subscript 0  space space space space space space Energi straight space potensialnya straight space colon  space space space space space space space space space EP equals mgh equals mg open parentheses 2 straight l minus 2 straight l space cos invisible function application space straight theta subscript 0 close parentheses  space space space space space space space space space EP equals 2 mgl open parentheses 1 minus cos invisible function application space straight theta subscript 0 close parentheses  space space space space space space space space space EP equals 2 mgl minus 2 mgl space cos straight space straight theta subscript 0  space space space space space space Maka comma straight space selisih straight space energi straight space di straight space titik straight space tengah straight space adalah straight space colon  space space space space space space space space space EK minus EP equals open parentheses 2 mgl minus 2 mgl space sin squared space straight theta subscript 0 close parentheses minus open parentheses 2 mgl minus 2 mgl space cos straight space straight theta subscript 0 close parentheses  space space space space space space space space space EK minus EP equals 2 mgl space cos invisible function application space straight theta subscript 0 minus 2 mgl space sin squared invisible function application space straight theta subscript 0  space space space space space space space space space EK minus EP equals 2 mgl left parenthesis cos invisible function application space straight theta subscript 0 minus sin squared invisible function application space straight theta subscript 0 right parenthesis  space space space space space space space space space EK minus EP equals 2 mgl left parenthesis cos space invisible function application straight theta subscript 0 minus open parentheses 1 minus cos squared invisible function application space straight theta close parentheses  space space space space space space space space space EK minus EP equals 2 mgl open parentheses cos invisible function application space straight theta subscript 0 plus cos squared invisible function application space straight theta subscript 0 minus 1 close parentheses  space space space space space space Masukan straight space straight theta subscript 0 equals 45 to the power of 0  space space space space space space space space space EK minus EP equals 2 mgl open parentheses 1 half square root of 2 plus open parentheses 1 half close parentheses minus 1 close parentheses  space space space space space space space space space EK minus EP equals 2 mgl open parentheses 1 half square root of 2 minus 1 half close parentheses  space space space space space space space space space EK minus EP equals 2 mgl open parentheses sin invisible function application space straight theta subscript 0 minus 1 half close parentheses  space space space space space space Hasil space ini space setara space dengan space colon  space space space space space space space space space EK minus EP equals 1 half left parenthesis 4 mgl right parenthesis open parentheses 2 space sin invisible function application space straight theta subscript 0 minus 1 close parentheses    left parenthesis 3 right parenthesis space Benar.  space space space space space space Energi space mekanik space bandul comma space yaitu space colon  space space space space space space space space space EM equals 1 half kA squared  space space space space space space space space space EM equals 2 mπ squared space straight f squared space straight A squared    left parenthesis 4 right parenthesis space Salah.  space space space space space space Perbandingan space periode space osilasi space di space bumi space dan space di space planet space tersebut space adalah space colon  space space space space space space space space space straight T subscript straight b over straight T subscript straight p equals fraction numerator 2 straight pi square root of straight L over straight g end root over denominator 2 straight pi square root of fraction numerator 2 straight L over denominator 2 straight g end fraction end root end fraction equals fraction numerator square root of 1 over denominator square root of 1 end fraction  space space space space space space Maka colon  space space space space space space space space space straight T subscript straight p equals straight space straight T subscript straight b    Jadi comma space hanya space pernyataan space left parenthesis 1 right parenthesis space dan space left parenthesis 3 right parenthesis space yang space benar. end style

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

5

Iklan

Iklan

Pertanyaan serupa

Sebuah bandul dengan panjang tali h diberi sudut simpangan awal θ 0 ​ sehingga berosilasi harmonik sederhana. Frekuensi di osilasinya adalah f 0 ​ . Percepatan gravitasi di permukaan bumi adalah g. Ji...

2

5.0

Jawaban terverifikasi

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

[email protected]

Contact 02140008000

02140008000

Ikuti Kami

©2024 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia