No negative sign on RHS and displacement appears as cubed, hence, not SHM

a = -10x follows the condition of SHM, acceleration\(\alpha \) -displacement hence, SHM.

A polyatomic molecule has a number of natural frequencies. So, in general, its vibration is superposition of SHMs of a number of different frequencies. Thus, superposition is periodic but not necessarily SHM.

A force which takes the body towards the mean postion in oscillation is called restoring force.
Characterstic of Restoring Force
The restoring force is aleays directed towards the mean positin and its magnitude of any instant is directly. Proportional to the displacement of the particle froom its mean postion of that instance.

The direction of acceleration of the bobat its mean position is radial i.e.towards the point of suspensiion.
At extreme points however, the acceleration is tagential towards the mean position.

Total distance travelled by an oscillator in one time period, from its mean position to one extreme position, then to other extreme position and finally back to mean position is 4 A, where A is the amplitude of oscillation.
Hence, the ratio =4A/A=4

When the displacement amplitude of the pendulum is extremely small as compared to its length.

According to force equation of SHM, F = - kx, where k is a constant known as force constant.

−0.4 m/s²

π rad,i.e.180^o

Given that,
Spring constant k =1200 Nm^-1
mass m=3 kg
The distance at which mass is pulled from its equilibrium x=0.2 cm
(i) Frequency,
\( f=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}} \)
\( =\frac{1}{2 \times 3.142} \times \sqrt{\frac{1200}{3}} \)
\( f=3.18 \mathrm{~Hz}\)
(ii) The acceleration is given by
\(a=-\omega^2 x=-\frac{k}{m} x\)
or \(\left|a_{\max }\right|=\frac{k}{m}\left|x_{\max }\right|\)
i.e., acceleration will be maximum, when x is maximum.
i.e., \(\mathrm{x}_{\max }=\mathrm{A}=0.02 \mathrm{~m}\)
\(\therefore \mathrm{a}=\frac{1200}{3} \times 0.02=8.0 \mathrm{~ms}^{-2}\)
(ii) The maximum speed of the mass is given by \( v=A \omega=A \sqrt{\frac{k}{m}}=0.02 \times \sqrt{\frac{1200}{3}} =0.40 \mathrm{~ms}^{-1} \)

When a ball is dropped from a height h, it gains velocity due to gravity pull. The body will enter the tunnel of earth with velocity, v =\(\sqrt{2gh}\)after a time, t =\(\sqrt{2h/g}\). The body will go out of earth on the other side through the same distance before coming back towards the earth. When the body is outside the earth, the restoring force F∝ (- 1/r²) and not (- r) so the motion does not remain SHM but becomes oscillatory. The period of oscillation of the body will be,
\(T=2\pi \sqrt { \frac { R }{ g } } +4\sqrt { \frac { 2h }{ g } } \) 

Area of cross-section of the block = \(\pi\)r² = 15 cm²
= 15 x 10^-4 m²
Total weight of the block = (230 + 50) = 280 gm = 0.28 kg
Density of wood = 0.30 gm/ c.c. = 300 kg/ m³
Density of water = 103 kg/ m³
When the cylinder is depressed in water through a distance y, the Restoring force weight of water displaced
F = Aydg = (15 x 10^-4) x 10³ x 9.8 Newton
Restoring force per unit distance= \(\frac{F}{y}\) = k
= (15 x 10^-4) x 10³ x 9.8 newton/ metre
= 1.5 x 9.8 N/m
Hence the frequency of oscillation is given by
\(=\frac { 1 }{ 2\pi } \sqrt { \left( \frac { k }{ m } \right) } =\frac { 1 }{ 2\pi } \sqrt { \frac { 1.5\times 9.8 }{ 0.28 } } \)
= 1.15 Hz

\(Here,m=10g,T=6s\)
\(\omega =\frac { 2\pi }{ T } =\frac { 2\pi }{ 6 } =\frac { \pi }{ 3 } rads^{ -1 }\)
\( When\ t=1s,v=6cms^{ -1 }\)
\(\\ As\ v=A\omega cos\omega t\)
\( 6=A\times \frac { \pi }{ 3 } cos\frac { \pi }{ 3 } \times 1=A\times \frac { \pi }{ 3 } cos{ 60 }^{ \circ }\)
\(=A\times \frac { \pi }{ 3 } \times \frac { 1 }{ 2 } =\frac { \pi A }{ 6 } orA=\frac { 36 }{ \pi } cm\)
\(Total\ energy,E=\frac { 1 }{ 2 } m{ A }^{ 2 }\omega ^{ 2 }\)
\(=\frac { 1 }{ 2 } \times 10\times \left( \frac { 36 }{ \pi } \right) ^{ 2 }\times \left( \frac { \pi }{ 3 } \right) ^{ 2 }=720erg\)
\(\\ Kinetic\ energy=\frac { 1 }{ 2 } m{ v }^{ 2 }=\frac { 1 }{ 2 } \times 10\times 6^{ 2 }=180erg\)
\(\\ \therefore Potential\ energy=Total\ energy-Kinetic\ energy\)
\( =720-180=540erg\)

(i) When t=0,x=0
⇒ 0=Asin(ωt+ϕ)
Asinϕ=0
or sinϕ=0 ∴ ϕ=0
∴ Required function is
x(t)=Asin(ωt+0) or x(t)=Asinωt
\(where,\ \omega =\sqrt { \frac { k }{ m } } =\sqrt { \frac { 1200 }{ 3.0 } } =20rad/s\)
∴ x(t)=Asin20t or x(t)=2sin20t
(ii) When t=0,x=+A
x(t)=Asin(ωt+ϕ) at t=0 and x=+A
+A=Asin(ω×0+ϕ)
or 1=sinϕ⇒ϕ=π/2
∴ x(t)=Asin(ωt+π/2)
=Acosωt=Acos20t=2cos20t
(iii) At t=0,x(t)=−A
⇒ −A=Asin(ω×0+ϕ) or −1=sinϕ or ϕ=3π/2
∴ x(t)=Asin(wt+3π/2)=−Acosωt=−2cos20t
So, the equations only differ in initial phase and in no other factors.

A second pendulum means a simple pendulum having time period T = 2s
For a simple pendulum T\(=2\pi \sqrt { \frac { l }{ g } } \)
where, l = length of the pendulum and g= acceleratin due to gravity On surface of the earth.
\({ T }_{ s }=2\pi \sqrt { \frac { l_{ e } }{ { g }_{ e } } } \)
On the surface of the moon,
\({ T }_{ n }=2\pi \sqrt { \frac { l_{ m } }{ { g }_{ m } } } \)
Dividing Eq(i) by Eq(ii) we get
\(\frac { { T }_{ s } }{ { T }_{ m } } =\frac { 2\pi }{ 2\pi } \sqrt { \frac { l_{ e } }{ { g }_{ e } } } \times \sqrt { \frac { l_{ m } }{ { g }_{ m } } } \)
T_s = T_m to maintain the second pendulum time period.
\(1=\sqrt { \frac { l_{ e } }{ { g }_{ e } } \times \frac { l_{ m } }{ { g }_{ m } } } \)
But the acceleration due to gravity at moon is 1/6 of the acceleration due to gravity at earth i.e.\({ g }_{ m }=\frac { { g }_{ e } }{ 6 } \) 
Squaring Eq (iii) and putting this value
\( 1=\frac { { l }_{ e } }{ { l }_{ m } } \times \frac { { g }_{ e }/6 }{ { g }_{ e } } =\frac { { l }_{ e } }{ { l }_{ m } } \times \frac { 1 }{ 6 } \)
\(\\ \Rightarrow \frac { { l }_{ e } }{ { 6l }_{ m } } =1\)
\(or { l }_{ m }=\frac { 1 }{ 6 } { l }_{ e }=\frac { 1 }{ 6 } \times 1=\frac { 1 }{ 6 } m\)