Two simple harmonic oscillators are on a horizontal frictionless table. All deta

Question

Two simple harmonic oscillators are on a horizontal frictionless table. All details about the oscillators and their oscillations are identical except their spring constants: kA and kB denote the spring constants for oscillators A and B, respectively and ka 2.kp How do the oscillator's maximum speeds compare? (a) vA.max-V2 - VB.max b) VBmax (c) vA,max-2 -VB.max (d) vB.max -2 VA.max Consider a tensioned string of length 1.4 m, linear mass density 0.70 g/m, and where disturbances travel at 45 m/s. If this string oscillates in its 2nd harmonic with amplitude 2.5 m, which of these is closest to the total energy stored in this standing wave? (be careful with this but you might find the expression for the total energy carried by one wavelength of a traveling sine wave useful: E- 1A2w2. Also, means 10-6)) (a) 0.19 - 10-9 J (b) 0.13 -109 J (c) 0.25 - 109J (d) 0.38 .10-9 J

Explanation / Answer

Solving 1st question of the page

We know maximum speed is

V=wA

Since everthing is same so A (amplitude is also same )

w=sqrt(k/m) so more is k ,more will be w and more speed

So speed ratio is

VA=sqrt(2)*VB

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