(i) Law of length:
For a given value of acceleration due to gravity, the time period of a simple pendulum is directly proportional to the square root of length of the pendulum.
T∝\(\sqrt{l}\)   .......(1)
(ii) Law of acceleration:
For a fixed length, the time period of a simple pendulum is inversely proportional to square root of acceleration due to gravity.
T∝\(\frac { 1 }{ \sqrt { g } } \)    .........(2)
 

Any oscillation driven by an external periodic agency to overcome the damping is known as forced oscillation.
Example: Sound boards of stringed instruments.

It is a special case of forced vibrations where the frequency of external periodic force (or driving force) matches with the natural frequency of the vibrating body (driven). As a result the oscillating body begins to vibrate such that its amplitude increases at each step and ultimately it has a large amplitude. Such a phenomenon is known as resonance and the corresponding vibrations are known is resonance vibrations.
Example: The breaking of glass due to sound.

When the oscillator is allowed to oscillate by displacing its position from equilibrium position, it oscillates with a frequency which is equal to the natural frequency of the oscillator. Such an oscillation or vibration is known as free oscillation or free vibration.

The effective value of acceleration due to gravity (g) will be equal to the component of g normal to the inclined plane which is \(g^{\prime}=g \cos \theta\)
\(T =2 \pi \sqrt{\frac{\ell}{g^{\prime}}}  \)
\(=2 \pi \sqrt{\frac{l}{g \cos \theta}}\)
Here length of pendulum l = 0.9 m
Angle of inclination \(\theta =45^{\circ}  \)
\(g^{\prime} =g \cos \theta  \)
\(=9.8 \cos 45^{\circ}=\frac{9.8}{1.414}  \)
\(=6.9306 \mathrm{~m} / \mathrm{s}^{2}\)
∴ Time period \(T =2 \pi \sqrt{\frac{\ell}{g \cos \theta}}  \)
\(=2 \times 3.14 \times \sqrt{\frac{0.9}{6.9306}}  \)
\(=6.28 \times \sqrt{0.129858}  \)
\(=6.28 \times 0.3602  \)
\(=2.262 \text { second }\)