Oscillations

In simple harmonic motion velocity
$ = A\omega \sin \left( {\omega t + \pi /2} \right)$



acceleration $ = A{\omega ^2}\sin \left( {\omega t + \pi } \right)$ from this we can easily find out that when v is maximum, then $a$ is zero.

Potential energy of particle performing SHM varies parabolically in such a way that at mean position it becomes zero and maximum at extreme position.

The resonance wave becomes very sharp when damping force is small.

For a simple harmonic motion between A and B, with O as the mean position, maximum kinetic energy of the particle executing SHM will be at O and maximum potential energy will be at A and B.


$ \therefore $ Displacement between maximum potential energy and maximum kinetic energy is $ \pm a$.

This is a case of damped vibration as the amplitude of vibration is decreasing with time.
Amplitude of vibrations at any instant t is given by $a = a_0e^{–bt}$, where $a_0$ is the initial amplitude of vibrations and b is the damping constant.

Now, when t = 100T, $a = a_0/3$  [T is time period]

Let the amplitude be $a'$ at t = 200T.
i.e. after completing 200 oscillations.

$ \therefore $ $a = a_0/3 = a0e^{–100Tb}$   ...(i)

and $a' = a_0e^{–200Tb}$ ...(ii)

From (i), ${1 \over 3} = {e^{ - 100Tb}}$
$ \therefore $ ${e^{ - 200Tb}} = 1/9$

From (ii), $a' = {a_0} \times {1 \over 9} = {{{a_0}} \over 9}$

$ \therefore $ The amplitude will be reduced to 1/9 of initial value.

The time period of a spring mass system as shown in figure 1 is given by $T = 2\pi \sqrt {m/k} $, where k is the spring constant.

$ \therefore $ ${t_1} = 2\pi \sqrt {m/{k_1}} $   ...(i)
and ${t_2} = 2\pi \sqrt {m/{k_2}} $   ...(ii)


Now, when they are connected in parallel as shown in figure 2(a), the system can be replaced by a single spring of spring constant, k_eff = k₁ + k₂.
[Since mg = k₁x + k₂x = k_effx]

$ \therefore $ ${t_0} = 2\pi \sqrt {m/{k_{eff}}} = 2\pi \sqrt {m/\left( {{k_1} + {k_2}} \right)} $   ...(iii)

From (i), ${1 \over {t_1^2}} = {1 \over {4{\pi ^2}}} \times {{{k_1}} \over m}$   ...(iv)

From (ii), ${1 \over {t_2^2}} = {1 \over {4{\pi ^2}}} \times {{{k_2}} \over m}$   ...(v)

From (ii), ${1 \over {t_0^2}} = {1 \over {4{\pi ^2}}} \times {{{k_1} + {k_2}} \over m}$   ...(vi)

Now (iv) + (v)

${1 \over {t_1^2}} + {1 \over {t_2^2}} = {1 \over {4{\pi ^2}m}}\left( {{k_1} + {k_2}} \right) = {1 \over {t_0^2}}$

$ \therefore $ $t_0^{ - 2} = t_1^{ - 2} + t_2^{ - 2}$

Energy = ${1 \over 2}m{\omega ^2}{a^2} = {1 \over 2}k{a^2}$

${f_A} = 2{f_B}$

$ \Rightarrow {1 \over {2\pi }}\sqrt {{g \over {{l_A}}}} = 2 \times {1 \over {2\pi }}\sqrt {{g \over {{l_B}}}} $

$ \Rightarrow {l \over {{l_A}}} = 4 \times {l \over {{l_B}}}$

$ \Rightarrow {l_A} = {{{l_B}} \over 4}$ , which does not depend on mass.

In $\Delta OAC,\,\cos \theta = A/l$

$ \Rightarrow OA = lcos\theta $

$ \therefore $ $AB = l\left( {1 - \cos \theta } \right) = h$

At point, C the velocity of bob = 0. The vertical acceleration = g

$ \therefore $ ${v^2} = 2gh$

$ \Rightarrow v = \sqrt {2gl\left( {1 - \cos \theta } \right)} $