Average kinetic energy and potential energy of a SHM
Total energy in SHM is given by, E=21mω2A2 where A is the amplitude and remains conserved. E=K+U Kavg=Uavg=2E=41mω2A2 Note: Average kinetic energy can also be found using Kavg=T1∫0TKdt Average potential energy can also be found using Uavg=T1∫0TUdt
Use the relation between restoring force and potential energy
Restoring force is given by: F=−dxdU It is often useful to find the equation of SHM. Example: A particle of mass 10 gm is placed in a potential field given by V=(50x2+100)J/kg. Find the frequency of oscillation in cycle/sec. Solution: Potential energy U=mV ⇒U=(50x2+100)10−2 F=−dxdU=−(100x)10−2 ⇒mω2x=−(100×10−2)x 10×10−3ω2x=100×10−2x ⇒ω2=100,ω=10 ⇒f=2πω=2π10=π5
Relation between restoring torque and potential energy
Restoring torque of a SHM can be found by: τ=−dθdU It is often useful in finding equation of SHM and helps in solving problems.
Potential energy per unit length at a given point in a travellling sound wave
Potential energy per unit length (w) at a point is defined as:
w=cpv p is the sound pressure. v is the particle velocity in the direction of propagation. c is the speed of sound.
Kinetic energy per unit length at a given point in a travelling sound wave
Kinetic Energy of a traveling sound wave is defined as: K.E.=∫2ρv2dV Example. Two pulses in a stretched string whose centres are initially 8 cm apart are moving towards each other as shown in figure. The speed of each pulse is 2 cm/s. After 2 seconds, what will be the total energy of the pulses.
Solution: After two seconds, the two pulses would nullify each other. As the string now becomes straight, there would be no deformation of the string. In such a situation, there would be no potential energy.
Total Potential Energy in one wavelength in a travelling sound wave
Total potential energy of a traveling sound wave is given by: P.E.=41μω2A2λ where, μ= Linear mass density, ω= Angular frequency, A= Amplitude of wave.
Derive and use kinetic energy per unit length at a given point in a travelling sound wave
Example: The velocity of a sound wave in v and the wave energy density is E, then find the amount of energy transferred per unit area per second by the wave in a direction normal to the wave propagation.
Solution: Energy transferred in normal direction to the wave propogation is also known as intensity I=21ρa2ω2v E=ω2a2ρ hence I=Ev