Time Period of Periodic and Simp


1	SHM	F=-kx or ma=-kx or a=-(k/m)x =-ω² x and T= 2π /ω = 2π√(m/k). We find m/k=x/a. Hence T=2π√(displacement/acceleration)
2	Combination of springs	For Series -- 1/k = 1/k1 + 1/k2 For parallel , k=k1+k2
3.	Simple pendulum with small angle of oscillation so that sinθ ≃ θ	mgcosθ =T, mgsinθ =-ma or gθ=-a or a=-gθ =-gx/l and T = 2π /ω = 2π√(l/g)
4.	A simple pendulum is hung from the roof of ceiling of lift by a mass less string of length l and executes shm	If the lift moves upward with uniform acceleration a, time period T= 2π√(l/{g+a}) If the lift moves downward with uniform acceleration a, time period T= 2π√(l/{g-a})
5.	A mass m hung from ceiling of a horizontally moving vehicle with acceleration a by rope of length l	T= 2π√(l/a') where a'² =g²+a² ;
6.	Torsion Pendulum hung with metallic wire of length l, having mass m, and radius r and moment of Inertia I = (1/2)mr² , angle of twist θ	Torque=τ =Iα =-cθ where c is the moment of force. α =-(c/I)θ , T= 2π /ω = 2π√(I/c) .Since c=F x 2r =2Fr, T=π√(mr/F) when r --> 0, F remaining constant, T --> 0 i.e. time speeds up. When r=0, there is no oscillation since it becomes a point mass. A point mass has no dimension since it does not occupy any space but at the same time , it has a spatial signature in the form of coordinates. Compare this with energy which has neither spatial signature nor dimension. This dichotomy in the treatment of point mass is reflected in quantum mechanical treatment where we refer to eigen states for both matter and energy.
7.	A mass m hung from ceiling by rope of length l moves making an angle θ in a circle of radius r on the table. What is T?	Tcosθ =mg , sinθ =r/l Tsinθ=mv²/r , time period t=2πr/v. Solving, we get t= 2π√(lcosθ/g)

8.	A mass M is hung from the hole in a table and a string of radius r with hole at the origin with mass m at the end executes a circle so that the M does not fall down	mv²/r=Mg or mω² r =Mg , ω =√(Mg/mr) , T=2π√(mr/Mg)
9.	A body of mass m executes a SHM in a straight line. If maximum velocity is a and maximum acceleration is b, then	displacement =y = a'sinωt, velocity=v=a'ωcosωt , acceleration=a= -a'ω²sinωt maximum velocity=a=a'ω . maximum acceleration=b=a'ω² . a/b=1/ω , T=2π /ω =2π a/b
10.	A body of mass m executes SHM. The displacement and velocity at 2 different points are v1,v2 and x1 and x2 respectively. A is amplitude	v1² = ω² (A²-x1²) , v2² = ω² (A²-x2²). Solvong, we get ω²=(v1²-v2²) /(x2²-x1²), T=2π /ω =2π √[(x2²-x1²)/(v1²-v2²) ]
11.	In uniform circular motion, if radius is a , speed is v (If speed will not be constant, then the acceleration will not be towards the center.)	F=GMm/a² =mv²/r and T=2πa /v and T² =4π²a² /v² .So T²=(4π²*/GM)a**³;
	In motion in an ellipse under the influence of a central force where the heavier mass M is at focus, and semi major axis is a	T²=(4π²*/GM)a**³
12	If a hole is drilled across the earth passing through the center and a mass of m is dropped into it, it will execute SHM.	Let the particle be at a distance x from center where x < R. F=-GMm/x² =-G(4/3)πx³dm/x² =ma a=-G(4/3)πxd , ω²=4Gπd/3 and T=2π√(3/4Gπd)
13	A sphere of mass m carrying positive charge q is hung from ceilingby a rope of length l amid a uniform electric field E directed vertically upward.	F=mg-qE=ma So a= g-(q/m)E and T=2π√(l/a)
14.	A point mass m is suspended at the end of a massless wire of length l and cross section A. If Y is Young's modulus of elasticity for the wire, rime period of oscillation along vertical line is T	Y =(F/A) / (Δl/l) =Fl/(AΔl) or F=YAΔl /l =-ma and so a=-(YA/ml)Δl , T= 2π /ω = 2π√(ml/YA)
15	A weighted glass tube is floating in a liquid with l of its length immersed. It is pushed down some distance x and released. It executes a SHM.	upward thrust=Aldg , At equilibrium, mg=Aldg or m=Ald F=Adg(l+x) -Adgl =adgx =-ma or a =-Adgx/Ald =-gx/l , T = 2π /ω = 2π√(l/g)
16.	A vertical U tube of uniform area of cross section a has water up to a height l. If water on one side is depressed and released, the motion is SHM.	m=A2ldg , F=-Ald2x =ma or a=-gx/l , T = 2π /ω = 2π√(l/g)
17.	The transverse vibration of a stretched string of length l tied to 2 ends creates a SHM.	T=(l/k)√(m/t) where m is mass per unit length, k is a number which for fundamental frequency is 1/2 and t is tension along the string. k=1/2, 1, 3/2, 2 etc
18.	The transverse vibration of a both side open organ pipe of length l creates a SHM.	T=(l/k)√(d/γp) where d is density , p is pressure and γ =Cp / Cv and k=1/2, 1, 3/2, 2 etc
	The transverse vibration of a one side open organ pipe of length l creates a SHM.(At closed end, node is formed)	T=(l/k)√(d/γp) and k=1/4, 3/4, 5/4, 7/4 etc
19.	Gas is enclosed in a container with area of cross section A, pressure P and volume V with a piston on the top.Piston is slightly pushed downward by x and released and it executes a SHM.	PV^γ=constant (P+dP)(V-dV)^γ =constant or PV^γ(1+dP/P)(1-dV/V)^γ =constant or (1+dP/P)(1-dV/V)^γ =1 Expanding the bold terms bionomially up to 2nd term, (1+dP/P)(1-γdV/V) =1 or dP/P=γdV/V or dP=γPdV/V *F=AdP =γPAdV/V =γPAAdx/V=γPA²dx/V =-ma o a =-dx( γPA²/mV) and T= 2π /ω = 2π√(mV/**γPA²).
20	In an LC circuit in oscillation of charge and current,	T =2π√(LC)
21	A pendulum of mass m with length l is suspended in a car that is travelling with a constant speed u around a circle of radius r. If the pendulum undergoes small oscillations, time period T is	T =T =2π√(l/a) where a² =g² -a'² where a' (relative acceleration)=v²/r and a'² =v⁴ /r²
22	For planetary motion around sun in elliptical orbits	T =2π√(a³/GM)
23	Two equal electric charges each q are placed at a distance 2l from each other and a charge of q is placed exactly in the middle. If it is displaced by a small distance x on either side, time period of oscillation	ΔF=F1-F2=kq²[1/(l-x)² - 1/(l+x)²] =kq²[ 4lx/(l²-x²)²]=4kq²lx /l⁴ if l >>x or ΔF=4kq²x /l³ ma=-(4kq²/l³)x or T=2π√(ml³/4kq²)
24	Charge q is uniformly distributed over a circular ring of radius r and the ring is held vertically. A opposite charge q0 is placed on the axis of the ring at a distance x from the center. If it is slightly disturbed along the axis, it executes SHM. Time period T is	dF=-kdqq0/a² where a²=r²+x² ; If a makes an angle A with the axis, dFsinA components cancel out and dFcosA=(kdqq0/a² )(x/a)=-(kdqq0/a³ )x F=ΣdFcosA=-(kqq0/a³ )x a³=(r²+x²)^3/2 If r>>x, a³=r³ ; w²=(kqq0/r³ ) and T=2π / w
25	A hollow sphere is filled with water and is used as pendulum bob.If water trickles out slowly through a hole made at the bottom, how will time period change?	T = 2π /ω = 2π√(l/g) Since T does not depend on mass, T will not change.