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Fundamentals Of Physics
Found in: Page 478

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Short Answer

In a demonstration, a 1.2 kg horizontal rope is fixed in place at its two ends (x = 0 and x = 2.0m) and made to oscillate up and down in the fundamental mode, at frequency 5.0 Hz. At t = 0, the point at x = 1.0m has zero displacement and is moving upward in the positive direction of a y axis with a transverse velocity of 5.0m/s. What are (a) the amplitude of the motion of that point and (b) the tension in the rope? (c) Write the standing wave equation for the fundamental mode.

  1. The amplitude of the motion of the point on the rope is 0.16m
  2. Tension in the rope is 2.4×102N
  3. The standing wave equation for the fundamental mode is,

y=(0.16m)sin1.57m-1xsin31.4radst

See the step by step solution

Step by Step Solution

Step 1: The given data

  1. Mass of the rope is, m = 1.2 kg
  2. Fundamental frequency on the rope is, f1=5.0 Hz
  3. At t = 0 , the point at x=1.0 m has zero displacement and velocity is, vm=5.0 m/s

Step 2: Significance of the wave equation

A wave is an oscillation that moves through space while also transferring energy. Without moving the particles of the medium, the waves' motion results in an energy transfer. We can find the amplitude of the motion of the point on the rope from its velocity and frequency using the corresponding relation. Using the formula for the velocity of particles on the string we can find tension in the rope. From values of angular velocity, wave number, and amplitude we can write the standing wave equation for the fundamental mode.

Formula:

The velocity of an oscillation, v=ωk …(i)

The maximum transverse speed, vm=ωym …(ii)

The angular frequency of the wave, ω=2πf …(iii)

The speed of the wave, v=fλ …(iv)

The velocity of the string, v=Tμ …(v)

The wavenumber of the wave, k=2πλ …(vi)

Step 3: (a) Determining the amplitude of the motion

From equation (iii), the angular velocity of the wave is given as:

ω=2π5 Hz =31.4rad/s

Hence, using value of angular frequency in equation (ii), we get the amplitude of the motion is given as:

ym=5m/s31.4rad/s =0.159m ~0.16m

Therefore, the amplitude of the motion of the point on the rope is 0.16 m

Step 4: (b) Determining the tension of the rope

Using equation (iv), the speed of waves on the rope is given as:

v=f2L λ=first fundamental wavelength=2L

The tension of the rope using equation (v) is given as:

T=v2mL μ=mL =2fL2mL =4f2Lm =45Hz2(2m)(1.2kg) =240 N =2.4×102N

Therefore, tension in the rope is 2.4×102N

Step 5: (c) Determining the equation of the standing wave

The wave number using equation (vi) is given as:

k=2(3.14)2(2m) λ=2L=1.57m-1

Since, at t = 0 , the point at x = 1.0 m has zero displacement, this is a sine wave. Its equation is given as:

y=ymsinkxsinωt=(0.16m)sin1.57m-1xsin31.4radst

Therefore, the standing wave equation for the fundamental mode is

y=(0.16m)sin1.57m-1xsin31.4radst

Most popular questions for Physics Textbooks

Body armor. When a high-speed projectile such as a bullet or bomb fragment strikes modern body armor, the fabric of the armor stops the projectile and prevents penetration by quickly spreading the projectile’s energy over a large area. This spreading is done by longitudinal and transverse pulses that move radially from the impact point, where the projectile pushes a cone-shaped dent into the fabric. The longitudinal pulse, racing along the fibers of the fabric at speed ahead of the denting, causes the fibers to thin and stretch, with material flowing radially inward into the dent. One such radial fiber is shown in Fig. 16-48a. Part of the projectile’s energy goes into this motion and stretching. The transverse pulse, moving at a slower speedvt, is due to the denting. As the projectile increases the dent’s depth, the dent increases in radius, causing the material in the fibers to move in the same direction as the projectile (perpendicular to the transverse pulse’s direction of travel). The rest of the projectile’s energy goes into this motion. All the energy that does not eventually go into permanently deforming the fibers ends up as thermal energy. Figure 16-48b is a graph of speed v versus time t for a bullet of mass 10.2g fired from a .38 Special revolver directly into body armor. The scales of the vertical and horizontal axes are set byvs=300m/sandts=4.00μs. TakevI=2000m/s, and assume that the half-angle θof the conical dent is60°. At the end of the collision, what are the radii of (a) the thinned region and (b) the dent (assuming that the person wearing the armor remains stationary)?

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