A block of mass attached to a spring is stretched and released from a distance of 7cm from the
equilibrium position. After 0.25 sec the instantaneous phase of the oscillation was π/3. What is the
angular frequency of the oscillation and write the equation of the SHO in exponential form?

We are tasked with determining the **angular frequency** \((\omega)\) and the **equation of motion** for the simple harmonic oscillator (SHO) in exponential form. Let’s break it down step by step.

---

### **1. Problem Analysis**
The given data are:
- Amplitude of oscillation: \(A = 7 \, \text{cm} = 0.07 \, \text{m}\)
- Instantaneous phase: \(\phi(t = 0.25 \, \text{s}) = \frac{\pi}{3}\)
- Time: \(t = 0.25 \, \text{s}\)

The general solution for the displacement \(x(t)\) in simple harmonic motion is:

\[
x(t) = A \cos(\omega t + \phi_0)
\]

Here:
- \(A\) is the amplitude.
- \(\omega\) is the angular frequency.
- \(\phi_0\) is the initial phase.

We aim to find \(\omega\) and write the solution in exponential form.

---

### **2. Relating Phase and Angular Frequency**
The instantaneous phase is given by:
\[
\text{Phase at } t = \omega t + \phi_0
\]

At \(t = 0.25 \, \text{s}\):
\[
\omega \cdot 0.25 + \phi_0 = \frac{\pi}{3}
\]

---

#### **Step 1: Initial Phase (\(\phi_0\))**
Assume the spring is stretched and released from rest at \(x = A\). This means the initial displacement \(x(0) = A\) and initial velocity \(v(0) = 0\). For such a condition, the initial phase is:
\[
\phi_0 = 0
\]

Thus, the phase relation simplifies to:
\[
\omega \cdot 0.25 = \frac{\pi}{3}
\]

---

#### **Step 2: Solve for \(\omega\)**
Rearranging for \(\omega\):
\[
\omega = \frac{\pi / 3}{0.25} = \frac{\pi}{0.75} \approx 4.19 \, \text{rad/s}
\]

---

### **3. Equation of Motion in Exponential Form**
The general exponential form of the SHO equation is:
\[
x(t) = A e^{i(\omega t + \phi_0)}
\]

Substituting the values:
- \(A = 0.07 \, \text{m}\)
- \(\omega \approx 4.19 \, \text{rad/s}\)
- \(\phi_0 = 0\)

The equation becomes:
\[
x(t) = 0.07 e^{i (4.19 t)}
\]

---

### **Final Answers**
1. **Angular frequency:** \(\omega \approx 4.19 \, \text{rad/s}\)
2. **Equation in exponential form:** \(x(t) = 0.07 e^{i (4.19 t)}\)

---

### Do you want a detailed explanation of any step?  
Here are 5 related questions:  
1. How do we interpret the instantaneous phase in SHO problems?  
2. Why does the initial phase \(\phi_0\) equal zero in this scenario?  
3. What is the physical meaning of the angular frequency?  
4. How can we derive the exponential form from the trigonometric form?  
5. Can you explain the relationship between angular frequency and period?  

**Tip:** Always check the units and assumptions (like initial conditions) when solving oscillation problems!

A block of mass attached to a spring is stretched and released from a distance of 7cm from the equilibrium position. After 0.25 sec the instantaneous phase of the oscillation was π/3. What is the angular frequency of the oscillation and write the equation of the SHO in exponential form?

Discover how to calculate the angular frequency and write the equation of a simple harmonic oscillator in exponential form. This problem involves key SHM concepts, angular frequency, and phase relations, suitable for Grades 11-12 students.

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