# Force, Transverse Traveling and Natural Frenquency of a System

4. A 70 kg ancient statue lies at the bottom of the sea (fresh water). Its volume is 3.0*10^4 (cm)^3.

a) How much force is required to lift it up at constant velocity through the water? (A force diagram would prove helpful in this problem).

b) If the statue was in salt water instead of fresh water, would the force required to lift the statue be greater or less than in fresh water? Explain.

5. A transverse traveling wave on a string is represented by

y(x,t) = 0.15 m sin [(0.393/m)x + (12.57rad/s)t]

where standard notation is used. For this wave, determine the

a) wavelength

b) frequency

c) velocity

d) amplitude

e) angular frequency and wave vector

f) magnitude of the maximum speed of particles on the string

g) magnitude of the maximum acceleration of particles on the string

h) direction of propagation of the wave

i) particle velocity and particle acceleration of the particle at x = 6m when t = 0.250s

6. a) Explain what characterizes a natural frequency of a system.

b) Explain what is meant by resonance, or driving the system at resonance.

c) Give an example of a system that has only one natural frequency and a system that has an infinite number of natural frequencies.

d) Explain what is meant by the principle of superposition.

https://brainmass.com/physics/resonance/force-transverse-traveling-natural-frenquency-system-348912

## SOLUTION This solution is **FREE** courtesy of BrainMass!

Please refer to the attachment.

Solution:

f

mg

a) This problem is solved using Archimedes' principle according to which an object immersed in a liquid experiences an upward force (thrust) which is equal to the weight of the liquid displaced by the body. Hence, the forces experienced by the statue are i) the weight mg acting downwards and ii) the Upward thrust f of sea water, as shown in the fig..

Net downward force acting on the statue = mg - f

Hence, a minimum force F = mg - f is required to lift the statue off the sea bed.

Volume of the water displaced by the statue = Volume of the statue = 3x104 cu.cm = 3x104x10-6 or 3x10-2 m3

Density of fresh water = 1000 kg/m3

Mass of the water displaced by the statue = 3x10-2x1000 = 30 kg

Weight of the water displaced by the statue = 30 x 9.8 = 294 N

Weight of the statue = mg = 70 x 9.8 = 686 N

Minimum force required to lift the statue = F = 686 - 294 = 392 N

b) As the density of salt water is slightly greater than the fresh water, the upward thrust f will also be slightly higher (weight of the statue remaining same). Hence, a little less force will be required to lift the statue.

Solution: Standard wave equation: y = Asin[(2ÐŸ/T)t + (2ÐŸ/Î»)x]

where: y(x,t) = Displacement of a particle on the string as a function of x and t

A = Amplitude of oscillation

Î» = Wave length

T = Time period

- sign indicates wave travelling in positive x direction (i.e towards right)

+ sign indicates wave travelling in negative x direction (i.e towards left)

Comparing the given equation with the standard equation we get the following:

a) 2ÐŸ/Î» = 0.393 Î» = 2ÐŸ/0.393 = 16 m

b) 2ÐŸ/T = 2ÐŸf = 12.57 f = 2 Hz

c) Velocity of the wave = Wave length x Frequency = 16 x 2 = 32 m/s

d) Amplitude = A = 0.15 m

e) Angular frequency Ï‰ = 2ÐŸf = 2 x 3.14 x 2 = 12.57 rad/sec

Wave vector k = 2ÐŸ/Î» = 2x3.14/16 = 0.39 per m

f) Speed of a given particle on the string = v = dy/dt = A(2ÐŸ/T)cos[(2ÐŸ/T)t + (2ÐŸ/Î»)x] .....(1)

Maximum magnitude of the speed = A(2ÐŸ/T) = A(2ÐŸf) = 0.15(2x3.14x2) = 1.88 m/s

g) Acceleration of a given particle on the string = a = d2y/dt2 = - A(2ÐŸ/T)2sin[(2ÐŸ/T)t + (2ÐŸ/Î»)x] ....(2)

Maximum magnitude of the acceleration = A(2ÐŸ/T)2 = A(2ÐŸf)2 = 0.15(2x3.14x2)2 = 23.66 m/s2

h) As the sign between the x term and t term is +ve, the wave is propagating in the -ve x direction i.e. towards the left.

i) Substituting values in equation (1), we get: v = 0.15x12.57cos[12.57t + (2ÐŸ/16)x]

v = 1.88cos[12.57t + 0.39x]

At x = 6m and t = 0.25 sec we have: v = 1.88cos[12.57(0.25) + 0.39(6)] = 1.87 m/s

For acceleration we have: a = - 0.15(12.57)2sin[12.57t + (2ÐŸ/16)x]

At x = 6m and t = 0.25 sec we have: a = - 23.7sin[12.57(0.25) + 0.39(6)] = - 2.26 m/s2

Solution: a) A mechanical system such as a mass suspended by a spring, a simple pendulum, a cantilever beam tend to vibrate at a fixed frequency (or frequencies) when disturbed from their mean (equilibrium) position. This frequency is known as the natural frequency of the system. The natural frequency depends upon the physical characterstics of the system. For example in case of a spring-mass system, it depends upon the spring constant (which in turn depends upon the material of the spring) and the mass.

b) If the driving force (force which creates the disturbance) acts on the system at the natural frequency of the system, the system tends to oscillate at the natural frequency but progressively increasing amplitudes (theoretically indefinitely). This is known as "driving the system at resonance".

c) A spring-mass system (ideally a point mass suspended vertically from one end of a mass less spring with the other end fastened to a rigid support) has only one natural frequency.

A string stretched between two rigid clamps (distributed mass) has an infinite number of natural frequencies known as the modes of vibration.

d) Given two waves superposing over each other, the resultant wave is given by the addition of the two superposing waves. This is known as superposition principle.

y1 = A1sin[(Ï‰1t + k1x]

y2 = A2sin[(Ï‰2t + k2x]

Then superposition of the above two waves gives the resultant wave:

y = A1sin[(Ï‰1t + k1x] + A2sin[(Ï‰2t + k2x]

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