Simple Harmonic Motion

Given: Two simple harmonic motions (SHMs) in the same direction with:
- Equal amplitude $A$
- Equal frequency
- Resultant amplitude = $A$
Formula for resultant amplitude when combining two SHMs:
$R%20%3D%20%5Csqrt%7BA%5E2%20%2B%20A%5E2%20%2B%202A%5E2%5Ccos%5Cdelta%7D%20%3D%20A%5Csqrt%7B2(1%20%2B%20%5Ccos%5Cdelta)%7D$
Given: $R%20%3D%20A$
So, $A%20%3D%20A%5Csqrt%7B2(1%20%2B%20%5Ccos%5Cdelta)%7D%20%5CRightarrow%20%5Csqrt%7B2(1%20%2B%20%5Ccos%5Cdelta)%7D%20%3D%201%20%5CRightarrow%202(1%20%2B%20%5Ccos%5Cdelta)%20%3D%201%20%5CRightarrow%20%5Ccos%5Cdelta%20%3D%20-%5Cfrac%7B1%7D%7B2%7D$
Therefore, $%5Cdelta%20%3D%20%5Ccos%5E%7B-1%7D%5Cleft(-%5Cfrac%7B1%7D%7B2%7D%5Cright)%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B3%7D$
Final Answer: $%5Cboxed%7B%5Cdelta%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B3%7D%7D$

Step 1: Understanding the SHM Equation The displacement equation for SHM is given as:

y%20%3D%20A%20%5Csin%20(%5Comega%20t%20%2B%20%5Cphi)

Where: -

y

is the displacement, -

A

is the amplitude, -

%5Comega

is the angular frequency, -

t

is the time, -

%5Cphi

is the phase constant. The given displacement equation is:

y%20%3D%202%20%5Csin%20%5Cleft(%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%2B%20%5Cphi%20%5Cright)%20%5C%2C%20%5Ctext%7Bcm%7D

Here, the amplitude

A%20%3D%202

cm, and the angular frequency

%5Comega

%5Cfrac%7B%5Cpi%7D%7B2%7D

. Step 2: Formula for Maximum Acceleration The acceleration in SHM is given by:

a%20%3D%20-%5Comega%5E2%20y

The maximum acceleration occurs when

y%20%3D%20A

, so:

a_%7B%5Ctext%7Bmax%7D%7D%20%3D%20%5Comega%5E2%20A

Step 3: Substituting the Known Values We know that: -

A%20%3D%202%20%5C%2C%20%5Ctext%7Bcm%7D

%5Comega%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%7D

Now, substituting these values into the formula for maximum acceleration:

a_%7B%5Ctext%7Bmax%7D%7D%20%3D%20%5Cleft(%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%5Cright)%5E2%20%5Ctimes%202

Simplifying:

a_%7B%5Ctext%7Bmax%7D%7D%20%3D%20%5Cfrac%7B%5Cpi%5E2%7D%7B4%7D%20%5Ctimes%202%20%3D%20%5Cfrac%7B%5Cpi%5E2%7D%7B2%7D%20%5C%2C%20%5Ctext%7Bcm%2Fsec%7D%5E2

Step 4: Conclusion The maximum acceleration of the particle is

%5Cfrac%7B%5Cpi%5E2%7D%7B2%7D%20%5C%2C%20%5Ctext%7Bcm%2Fsec%7D%5E2

, which corresponds to:

%5Cboxed%7B(A)%7D%20%5C%2C%20%5Cfrac%7B%5Cpi%5E2%7D%7B2%7D%20%5C%2C%20%5Ctext%7Bcm%2Fsec%7D%5E2

Chapter Questions
3 MCQs