As mentioned in one of our previous posts [ Brainwave series ] simple devices could be used to determine important physical science functions as well as applications. The Pendulum is one such device . It consists of a rigid body mounted on a fixed horizontal axis, about which it is free to rotate under the influence of gravity. The period of the motion of a pendulum is quite independent of its amplitude and depends basically on a ) the geometry of the pendulum and on b) the local value of g, the acceleration of gravity. Pendulums are hence used as the control elements in clocks, or inversely as instruments to measure g.Pendulum – general representation
In the general representation of the pendulum shown above , O is the axis and C represents the center of mass. The line OC makes an instantaneous angle θ with the vertical. Considering the rotary motion of any rigid body about a fixed axis, the angular acceleration is equal to the torque about the axis divided by the moment of inertia I about the axis. Also , m represents the mass of the pendulum, and the force of gravity can be considered as the weight mg acting at the center of mass C.
Animation : the velocity and acceleration vectors of a simple pendulum.
- The motion is simple harmonic when the amplitude of motion of the pendulum is small . Period T, i.e time taken for a complete oscillation is given by 2π √ ( I / mgh )
- In the case of a simple pendulum the lengths h and L become identical, and therefore , the moment of inertia I equals mL2 . Period T becomes 2π √ L/g Note that the period is independent of the mass of the pendulum.
- If the amplitude is large , 2π √ [ l/g ( 1 + A²/16 ) ] where A is the amplitude expressed in radians.
- If the length of the pendulum is large , g becomes directed towards the centre of the earth and T = 2π √ [ 1/g ( 1/ l + 1 / R ) ]
- If l << radius of the earth ( R ) , then T = 2π √ l/g
- If the length of the pendulum is at infinity , then T = 2π √ R/g
- If the simple pendulum is in a vehicle accelerating up with acceleration |a| , then T = 2π √ [ l/(g + a) ]
- If the simple pendulum is in a vehicle accelerating down with acceleration |a| , then T = 2π √ [ l/(g – a) ]
- If the simple pendulum is in a vehicle accelerating in a horizontal direction with acceleration |a| , then T = 2π √ [ l/(g² + a² ) ]
- If the simple pendulum is in a falling lift , then g = 0 and T = ∞ ( no oscillation )
Now on to some questions –
1.The period of oscillation of simple pendulum of length L suspended from the roof of the vehicle which moves without friction , down an inclined plane of inclination α is given by – [ IIT JEE 2000 ]
- (a) 2π √ L/g cos α
- (b) 2π √ L/g sin α
- (c) 2π √ L/g
- (d) 2π √ L/g tan α
Ans – This case is a variation of no.8 mentioned above [ geff = g – a ] , where effective g gets resolved to sin and cos components. Now , if α is the angle of inclination , g ( acceleration of the bob ) becomes g cos α and acceleration of the point of suspension is g sin α .
Hence , geff = g – a = √ [ g² + (g sin α )² + 2 ( g ) ( g sin α ) cos ( 90° + α ) ] = g cos α . Hence , answer is option (a).
2.A simple pendulum has time period T1. The point of suspension is now moved upward according to the relation y = kt² , ( k = 1m/s²) where y is the verical displacement . The time period now becomes T2. The ratio of T1²/ T2² is ( Take g = 10 m/s² ) [ IIT JEE 2005 ]
- (a) 6/5
- (b) 5/6
- (c) 1
- (d) 4/5
Ans y = kt² , d²y/dt² = 2k . As k = 1m/s², ay = 2 m/s² .
T1 = 2π √ l/g and T2 = 2π √ L/g +ay , therefore , T1²/ T2² = 10 + 2/10 = 6/5 . So , correct option is (a).
3. The time period of a second’s pendulum at a depth R/2 from the earth’s surface is given by –
- (a) 3 √2 s
- (b) 5 √2 s
- (c) 2 √2 s
- (d) 4 √2 s
At a depth h below the earth’s surface , geff = g (1 – h/R) . Hence T2/T1 = √ g1/g2 = √ R / (R-h)
T2 = T1 √ R/R-h , hence T2 α 1/ √ (R-h) , therefore, T R/2 = 2 √ R/( R – R/2 ) = 2√2 s . Hence correct option is (c).
4. The bob of a simple pendulum ( density σ ) is immersed in a liquid of density ρ . The length of the string of the pendulum is l . Then the angular frequency of the small oscillations of the bob is –
- (a) T = 2π √ [ l/( g – Vρg/m) ]
- (b) T = 2π √ [ l/( g + Vρg/m) ]
- (c) T = 2π √ [ g – Vρg/m l ]
- (d) T = 2π √ [ g + Vρg/m l ]
Ans . The key factor here is that the bob experiences buoyant forces due to the liquid in addition to the usual gravitational and tensional forces . Now , the buoyant force = Vρg and so the net force on the bob = F = mg – Vρg . As the bob is undergoing small oscillations , its getting displaced and consequently there is a restoring force. Let the displacement be x ,
F = ( mg – Vρg ) sin θ = – ( mg – Vρg ) x/l . Now , acceleration = – ( g – Vρg /m ) x/l .
The standard form of SHM is a = – ω² x and so ω = √ [ ( g – Vρg /m )/ l ] .
Hence , T = 2π √ [ l/( g – Vρg/m) ] . option (a) is correct.