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1. How many beats can be heard by sounding together two tuning forks of frequencies 250Hz and 256Hz?
Number of beats (n) =f2- f1
= 256 - 250
= 6 beats /sec
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Not even a single member ofAksionov'sfamily was concerned about his well-being when he was at his lowest.Along with the judges of the court even his wife did not trust him.Hewas imprisoned for the crime he did not commit.With time he learnt to forgive people, but he did not forget their mistreatment.
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Here, f(x)=x2 -6 logx-3=0
f(2)=4-6 log2-3=-0.806
f(3)=9-6 log3-3=3.1373
f(2).f(3)=-0.806*3.1373=-2.529422 which is negative.
Hence, the root lies between 2 and 3
c0 =(2+3)/2=2.5
f(2.5)=6.25-6 log 2.5-3=0.8623
Now
n | a(-ve) | b(+ve) | cn | f(cn) |
0 | 2 | 3 | 2.5 | 0.8623 |
1 | 2 | 2.5 | 2.25 | -0.050595 |
2 | 2.25 | 2.5 | 2.375 | 0.38664 |
3 | 2.25 | 2.375 | 2.3125 | 0.1631658 |
4 | 2.25 | 2.3125 | 2.28125 | 0.05506 |
5 | 2.25 | 2.28125 | 2.265625 | 0.001925 |
From the table,
f(2.265625)=0.001928<10-2
Therefore, the...
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