To solve the problem, we need to evaluate the three statements provided regarding simple harmonic motion (SHM) and determine their validity.
### Step 1: Analyze Statement 1
**Statement 1:** In SHM, the scalar product of velocity and acceleration is always negative.
- The displacement in SHM can be expressed as \( x(t) = A \sin(\omega t) \).
- The velocity \( v(t) \) is the first derivative of displacement:
\[
v(t) = \frac{dx}{dt} = A \omega \cos(\omega t)
\]
- The acceleration \( a(t) \) is the second derivative of displacement:
\[
a(t) = \frac{d^2x}{dt^2} = -A \omega^2 \sin(\omega t)
\]
- The scalar product (dot product) of velocity and acceleration is given by:
\[
\mathbf{a} \cdot \mathbf{v} = |\mathbf{a}| |\mathbf{v}| \cos(\theta)
\]
where \( \theta \) is the angle between the velocity and acceleration vectors. In SHM, velocity and acceleration are always in opposite directions, meaning \( \theta = 180^\circ \).
- Therefore, \( \cos(180^\circ) = -1 \), leading to:
\[
\mathbf{a} \cdot \mathbf{v} = -|\mathbf{a}| |\mathbf{v}|
\]
- This confirms that the scalar product is always negative.
**Conclusion for Statement 1:** True.
### Step 2: Analyze Statement 2
**Statement 2:** The time period of a simple pendulum of very large length compared to Earth's radius is 1.4 hours.
- The formula for the time period \( T \) of a simple pendulum is:
\[
T = 2\pi \sqrt{\frac{L}{g}}
\]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.
- For very large lengths (i.e., \( L \to \infty \)), the time period approaches:
\[
T = 2\pi \sqrt{\frac{L}{g}} \to \infty
\]
- However, if we consider a specific case where \( L \) is very large but finite, we can calculate it. For a pendulum with a length comparable to the radius of the Earth, we can derive:
\[
T = 2\pi \sqrt{\frac{R_e}{g}} \approx 2\pi \sqrt{\frac{6400 \times 10^3 \, \text{m}}{9.8 \, \text{m/s}^2}} \approx 84.2 \, \text{minutes} \approx 1.4 \, \text{hours}
\]
- This confirms that the statement is indeed true.
**Conclusion for Statement 2:** True.
### Step 3: Analyze Statement 3
**Statement 3:** For a given amplitude of SHM, the total energy of a spring-mass system is independent of the mass of the body.
- The total energy \( E \) of a spring-mass system in SHM is given by:
\[
E = \frac{1}{2} k A^2
\]
where \( k \) is the spring constant and \( A \) is the amplitude.
- Notably, this expression does not contain the mass \( m \) of the body. The energy depends only on the spring constant and the amplitude.
- Thus, for a given amplitude, the total energy is indeed independent of the mass of the body.
**Conclusion for Statement 3:** True.
### Final Conclusion
All three statements are true.
