Roboguru

Benda bermassa 40 gram sedang bergerak harmonik sederhana dengan amplitudo 2,0 cm. Percepatan getar maksimumnya adalah .  Ketika simpangan benda 1/4 amplitudo besar energi kinetik benda tersebut adalah ...

Pertanyaan

Benda bermassa 40 gram sedang bergerak harmonik sederhana dengan amplitudo 2,0 cm. Percepatan getar maksimumnya adalah 3 comma 2 cross times 10 cubed space straight m divided by straight s squared.  Ketika simpangan benda 1/4 amplitudo besar energi kinetik benda tersebut adalah ... space

Pembahasan Soal:

Diketahui. 
m = 40 g = 0,04 kg
A = 2 cm = 0,02 m
a subscript m equals space 3 comma 2 space cross times 10 cubed space straight m divided by straight s squared y space equals space 1 fourth straight A space equals space 1 fourth left parenthesis 0 comma 02 right parenthesis equals 0 comma 005 space straight m

Maka, Energi kinetik pada simpangan 1/4 amplititudonya adalah.

Langkah 1. Mencari bold italic omega dari persaamaan percepatan maksimum

table attributes columnalign right center left columnspacing 0px end attributes row cell a subscript m a k s end subscript end cell equals cell space omega squared space A end cell row cell 3 comma 2 space cross times 10 cubed end cell equals cell space omega squared space space left parenthesis 2 cross times 10 to the power of negative 2 end exponent right parenthesis end cell row cell omega squared space end cell equals cell fraction numerator 3 comma 2 space cross times 10 cubed over denominator 2 cross times 10 to the power of negative 2 end exponent end fraction end cell row cell omega squared space end cell equals cell 1 comma 6 space cross times 10 to the power of 5 end cell row cell omega space end cell equals cell square root of 16 space cross times 10 to the power of 4 end root end cell row cell omega space end cell equals cell 4 space cross times 10 squared end cell row cell omega space end cell equals cell equals space 400 space Rad divided by straight s end cell end table

Langkah 2. Mencari Energi kinetik dengan bold italic v bold equals bold italic omega square root of bold A to the power of bold 2 bold minus bold y to the power of bold 2 end root

table attributes columnalign right center left columnspacing 0px end attributes row cell E k space end cell equals cell space 1 half m space v squared end cell row blank equals cell space 1 half m space left parenthesis omega square root of A squared minus y squared end root right parenthesis squared end cell row blank equals cell space 1 half m space open parentheses omega square root of A squared minus open parentheses 1 fourth A close parentheses squared end root close parentheses squared end cell row blank equals cell 1 half m space open parentheses omega square root of A squared minus 1 over 16 A squared end root close parentheses squared end cell row blank equals cell 1 half m space open parentheses omega square root of 15 over 16 A squared end root close parentheses squared end cell row blank equals cell 1 half left parenthesis 0 comma 04 right parenthesis space open parentheses 400 square root of 15 over 16 left parenthesis 0 comma 02 right parenthesis squared end root close parentheses squared end cell row blank equals cell 1 half left parenthesis 0 comma 04 right parenthesis space open parentheses 400 square root of 0 comma 000375 end root close parentheses squared end cell row blank equals cell left parenthesis 0 comma 02 right parenthesis left parenthesis 400 squared left parenthesis 0 comma 000375 right parenthesis right parenthesis end cell row blank equals cell space left parenthesis 0 comma 02 right parenthesis left parenthesis 60 right parenthesis end cell row blank equals cell space 1 comma 2 space straight J end cell end table

Oleh karena itu, jawabannya adalah 1,2 J. space

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 19 Mei 2021

Ruangguru

RUANGGURU HQ

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

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved