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Figure (a)is a partial graph of the position function x(t)for a simple harmonic oscillator with an angular frequency of 1.20鈥塺补诲/s ; Figure (b) is a partial graph of the corresponding velocity function v(t). The vertical axis scales are set by xs=5.0cm and vs=5.0鈥塩尘/s. What is the phase constant of the SHM if the position function x(t)is in the general form x=xmcos(t+)?

Short Answer

Expert verified

The phase constant of the SHM, if the position function x(t)is in the formx(t)=xmcos(t+f) , is 0.695鈥塺补诲.

Step by step solution

01

Stating the given data

  1. Angular frequency of the harmonic oscillator, =1.20鈥塺补诲/s
  2. Vertical axis scale values,xs=5.0鈥塩尘/s and.vs=5.0鈥塩尘/s
02

Understanding the concept of simple harmonic motion

Using the formula of velocity function and position function, we can find the phase constant of SHM by taking the ratio of velocity and position functions.

Formulae:

The general expression for velocity of motion,x=xmcos(t+f) (i)

The general expression for velocity of motion, v=xmsin(t+f) (ii)

03

Calculation of phase constant

Dividing equations(ii) by (i), we get

v(t)x(t)=xmsin(t+f)xmcos(t+f)

At
t=0鈥塻,v0=5鈥塩尘/s,x0=5鈥塩尘

v0x0=sin(f)cos(f)v0x0=tanff=tan1v0x0=tan1[(5鈥塩尘/s)(1.20rad/s)(5鈥塩尘)]=0.695鈥塺补诲

Therefore,the phase constant of the SHM, if the position functionsx(t)is in the form

x(t)=xmcos(t+f), is 0.695鈥塺补诲.

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Most popular questions from this chapter

You are to complete Figure 15-22a so that it is a plot of velocity v versus time t for the spring鈥揵lock oscillator that is shown in Figure 15-22b for t=0. (a) In Fig.15-22a, at which lettered point or in what region between the points should the (vertical) v axis intersect the t axis? (For example, should it intersect at point A, or maybe in the region between points A and B?) (b) If the block鈥檚 velocity is given byv=-vmsin(t+), what is the value of? Make it positive, and if you cannot specify the value (such as+/2rad), then give a range of values (such as between 0 and/2rad).

In Figure 15-41, block 2 of massoscillates on the end of a spring in SHM with a period of20ms.The block鈥檚 position is given byx=(1.0cm)cos(蝇t+/2)Block 1 of mass4.0kgslides toward block 2with a velocity of magnitude6.0m/s, directed along the spring鈥檚 length. The two blocks undergo a completely inelastic collision at timet=5.0ms. (The duration of the collision is much less than the period of motion.) What is the amplitude of the SHM after the collision?

When the displacement in SHM is one-half the amplitude Xm,

  1. What fraction of the total energy is kinetic energy?
  2. What fraction of the total energy is potential energy?
  3. At what displacement, in terms of the amplitude, is the energy of the system half kinetic energy and half potential energy?

The suspension system of a2000kgautomobile 鈥渟ags鈥10cmwhen the chassis is placed on it. Also, the oscillation amplitude decreases by 50% each cycle.

  1. Estimate the value of the spring constant K.
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The balance wheel of an old-fashioned watch oscillates with angular amplituderadand periodrole="math" localid="1655102340305" 0.500s.

  1. Find the maximum angular speed of the wheel.
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