Which of the following equation does not represent a simple harmonic motion ?

Q: Which of the following equation does not represent a simple harmonic motion ?

Option 4, $ y = b \tan \omega t $, does not represent a simple harmonic motion (SHM). Here's why: 1. **Explanation for Option 4:** - Simple harmonic motion is characterized by oscillatory motion that can be described using sine or cosine functions. These functions are periodic and bounded, meaning they repeat at regular intervals and have a maximum and minimum value. - The function $ y = b \tan \omega t $ involves the tangent function, which is not periodic in the same way as sine and cosine. The tangent function has vertical asymptotes and is undefined at odd multiples of $\frac{\pi}{2}$. This results in a graph that is not bounded and does not repeat in a regular, predictable pattern like sine or cosine functions. Therefore, it cannot represent SHM. 2. **Explanation for Other Options:** - **Option 1: $ y = a \sin \omega t + b \cos \omega t $** - This equation is a linear combination of sine and cosine functions, both of which are characteristic of SHM. The resulting function is still periodic and bounded, making it a valid representation of SHM. - **Option 2: $ y = a e^{i\omega t} $** - This expression represents a complex exponential function, which can be related to SHM through Euler's formula: $ e^{i\omega t} = \cos \omega t + i \sin \omega t $. The real part, $ \cos \omega t $, and the imaginary part, $ \sin \omega t $, both describe SHM. Thus, this option can represent SHM when considering the real or imaginary components. - **Option 3: $ y = b \cos \omega t $** - This is a standard form of SHM, as it directly uses the cosine function, which is periodic and bounded. It perfectly describes the oscillatory nature of SHM. In conclusion, Option 4 is the correct answer because the tangent function does not exhibit the periodic and bounded characteristics necessary for simple harmonic motion.

Question

Accepted Answer

Option 4, $ y = b \tan \omega t $, does not represent a simple harmonic motion (SHM). Here's why:

1. **Explanation for Option 4:**
   - Simple harmonic motion is characterized by oscillatory motion that can be described using sine or cosine functions. These functions are periodic and bounded, meaning they repeat at regular intervals and have a maximum and minimum value.
   - The function $ y = b \tan \omega t $ involves the tangent function, which is not periodic in the same way as sine and cosine. The tangent function has vertical asymptotes and is undefined at odd multiples of $\frac{\pi}{2}$. This results in a graph that is not bounded and does not repeat in a regular, predictable pattern like sine or cosine functions. Therefore, it cannot represent SHM.

2. **Explanation for Other Options:**
   - **Option 1: $ y = a \sin \omega t + b \cos \omega t $**
     - This equation is a linear combination of sine and cosine functions, both of which are characteristic of SHM. The resulting function is still periodic and bounded, making it a valid representation of SHM.
   
   - **Option 2: $ y = a e^{i\omega t} $**
     - This expression represents a complex exponential function, which can be related to SHM through Euler's formula: $ e^{i\omega t} = \cos \omega t + i \sin \omega t $. The real part, $ \cos \omega t $, and the imaginary part, $ \sin \omega t $, both describe SHM. Thus, this option can represent SHM when considering the real or imaginary components.
   
   - **Option 3: $ y = b \cos \omega t $**
     - This is a standard form of SHM, as it directly uses the cosine function, which is periodic and bounded. It perfectly describes the oscillatory nature of SHM.

In conclusion, Option 4 is the correct answer because the tangent function does not exhibit the periodic and bounded characteristics necessary for simple harmonic motion.