# The amplitude of SHM y=2(sin5πt+√2cosπt) is

## Question

The amplitude of SHM y=2(sin5πt+√2cosπt) is

The final answer to the amplitude of SHM is 2√3

### Explanation

From the equation of simple harmonic motion which says

y = A1sinωt + A2cosωt

We will now look into the equation from the question where

y = 2(sin5πt+√2cosπt)

By expanding the bracket by the right hand side, we will have

y = 2 x sin5πt + 2 x √2cosπt

The above expression will now become

y = 2sin5πt + 2√2cosπt

If we look at the equation y = A1sinωt + A2cosωt and compare it with the above equation, we will see that

y = 2sin5πt + 2√2cosπt

and

y = A1sinωt + A2cosωt

We can easily compare A1 with 2

And compare A2 with 2√2

Hence, we can say that

A1 = 2

Also

A2 = 2√2

Therefore, we can now calculate the total amplitude by applying the formula

A = √(A12 + A22)

By inserting the values of A1 and A2 into the above equation, we will now have

A = √(22 + (2√2)2) = (4 + 8)

Hence, the above expression will now become

A = 12 = (4 x 3) = 4 x 3 = 23

Therefore, the amplitude of the simple harmonic motion SHM is 2√3

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## Source

National institute of open schooling