Three strings of a musical instrument vibrate six, eight and 12 times the second respectively. If all the three begin to vibrate simultaneously, then find the sort is time interval before all three vibrate together again?
Answers
LCM OF 6,8&12 = 24 sec
The shortest time interval before all three vibrate together again is 24 seconds
Solution:
Given that,
Three strings of a musical instrument vibrate 6 , 8 and 12 times the second respectively
All the three begin to vibrate simultaneously, then the shortest time interval before all three vibrate together again is least common multiple of 6 , 8 and 12
Find LCM of 6 , 8 and 12
List all prime factors for each number.
Prime Factorization of 6 is: 2 x 3
Prime Factorization of 8 is: 2 x 2 x 2
Prime Factorization of 12 is: 2 x 2 x 3
For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.
2, 2, 2, 3
Multiply these factors together to find the LCM.
LCM = 2 x 2 x 2 x 3 = 24
Thus the shortest time interval before all three vibrate together again is 24 seconds
Learn more:
Four bells are heard at intervals of 2,3,5 and 7 minutes respectively. If all bells toll together exactly at 9 a.M., they will again be heard together at
https://brainly.in/question/9290719
Four bells toll at an interval of 8, 12 , 15 and 18 seconds respectively. All the four begin to toll together. How many times will they
https://brainly.in/question/4100885