The amplitude is A = Aoe^-\(\alpha\)t
Divide by Ao' take natural logarithms and solve for \(\alpha\)
\(\frac { A }{ { A }_{ 0 } } ={ e }^{ -\alpha t }\Rightarrow { l }_{ n }\left( \frac { A }{ { A }_{ 0 } } \right) =-\alpha t\)
\(\Rightarrow =-\frac { 1 }{ t } { l }_{ n }\left( \frac { A }{ { A }_{ 0 } } \right) \)
Since \(\alpha =\frac { \beta }{ 2m } \), the\(\beta\) =2m\(\alpha\).Thus
\(\beta =-\frac { 2m }{ t } { l }_{ n }\left( \frac { A }{ { A }_{ 0 } } \right) \)
Now substituting 0.950 kg for m, 600 s, for t, and 0.10 for\(( \frac { A }{ { A }_{ 0 } } )\)
\(\beta =-\frac { 2m }{ t } { l }_{ n }\left( \frac { A }{ { A }_{ 0 } } \right) =-\frac { 2(0.950kg) }{ 600s } { l }_{ n }\)0.10
= 7.3\(\times\)10^-3kg / s.

On earth time period
T = 2\(\pi \sqrt { \frac { l }{ g } } \)
3.5 = 2\(\pi \sqrt { \frac { l }{ 9.8 } } \)
\(\therefore\) Acceleration due to gravity of the moon's is 1.7 ms^-2
T¹ = \(\sqrt { \frac { l }{ 1.7 } } \)
Dividing \(\frac { { T }^{ 1 } }{ 3.5 } =2\pi \sqrt { \frac { 1 }{ 1.T } } /2\pi \sqrt { \frac { l }{ 9.8 } } v\)
T¹ = \(\sqrt { \frac { 9.8 }{ 1.7 } } \times 3.5\)
= 8.4 s

We know that T = 2\(\pi \sqrt { \frac { M }{ k } } \) ,where k=spring constant
In first case, 2=2\(\pi \sqrt { \frac { M }{ k } } \) ----------(1)
In second case, 3= 2\(\pi \sqrt { \frac { M+2}{ k } } \) ----------(2)
Squaring equation (1) and (2) and then dividing (2) by (1), we have,
\(\frac { { 3 }^{ 2 } }{ { 2 }^{ 2 } } =\frac { M+2 }{ M } \)
\(\frac { 9 }{ 4 } =\frac { M+2 }{ M } =1+\frac { 2 }{ M } \)
\(\frac { 9 }{ 4 } =1+\frac { 2 }{ M } \)
Solving we get, initial mass M = 1.6 kg.

x = a sin wt + β cos wt
α = r cos θ and, B = r sin θ
x = r cos θ sin wt + r sin θ cos
wt = r sin (wt + θ)

(i) When a pendulum clock gains time, it means it has gone fast, it makes more vibrations per day them required.
(ii) This shows that the time period of oscillations has decreased. Therefore, to correct it, the length of pendulum should be properly increased.

(i) When amplitudes of the vibrating pendulum is small then pendulum is small. Here the restoring force F = mg sin θ = mg θ = mgx/l .
(ii) Where x is the displacement of the bob and I is the length of pendulum. Hence F x x. Since F is directed towards mean position.
(ill) Therefore the motion of the bob of simple pendulum will be S.H.M. if θ is small.

When an object or a particle moves back and forth repeatedly for some duration of time its motion is said to be oscillatory (or vibratory).
Eamples: our heart beat, swinging motion of the wings of an insect,

The displacement time t = 0s(initial time), the phase φ=φ0 is called epoch (initial phase) where φ0 is called the angle of epoch.

Case(a):
Pulley is fixed rigidly here.
When the mass displace by y and the spring will also stretch by y.
Hence, F = T = ky
\(T=2\pi \sqrt { \frac { m }{ k } } \)
Case(b):
Mass displace by y, pulley also displaces by y.
T = 4ky.
\(T=2\pi \sqrt { \frac { m }{ 4k } } \)

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

Frequency of simple harmonic motion is defined as the number of oscillations produced by the Particle.

Time period is defined as the time taken by a particle to complete one oscillation in simple harmonic motion.
T = \(\frac { 2\pi }{ \omega } \)

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.