## Question

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

## Answer

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

### Explanation

From the equation of simple harmonic motion which says

y = A_{1}sinωt + A_{2}cosωt

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 = A_{1}sinωt + A_{2}cosωt and compare it with the above equation, we will see that

y = **2**sin5πt + **2√2**cosπt

and

y = **A _{1}**sinωt +

**A**cosωt

_{2}We can easily compare **A _{1}** with

**2**

And compare **A _{2} **with

**2√2**

Hence, we can say that

**A _{1}** =

**2**

Also

**A _{2} **=

**2√2**

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

**A = √(A _{1}^{2} + A_{2}^{2})**

By inserting the values of **A _{1} and A_{2}** into the above equation, we will now have

**A = √(2 ^{2} + (2√2**)

**=**

^{2})**√**(4 + 8)

Hence, the above expression will now become

A = **√**12 = **√** (4 x 3) = **√**4 x **√**3 = 2**√**3

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

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