A Piano Wire With Mass 300 G And Length 800 Cm Is Stretched

A Piano Wire With Mass 300 G And Length 800 Cm Is Stretched With A T

A piano wire with mass 3.00 g and length 80.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplitude 1.6 mm travels along the wire. (a) Calculate the average power carried by the wave. (b) What happens to the average power if the wave amplitude is halved?

Paper For Above instruction

In this paper, we analyze the physical properties of a vibrating piano wire, focusing on calculating the average power transmitted by a wave traveling along the wire. Specifically, given the mass, length, tension, frequency, and amplitude of the wave, we will determine the power transmitted, highlighting its dependence on wave parameters such as amplitude and frequency. Furthermore, we explore how halving the wave amplitude affects the average power carried by the wave.

Introduction

Waves on a string or wire are fundamental phenomena in physics, with applications ranging from musical instruments to engineering systems. The energy transmitted by a wave, typically quantified as power, depends on the wave's physical properties, including amplitude, frequency, tension, and mass per unit length. Understanding these relationships is crucial for designing systems that harness wave transmission efficiently, such as musical instruments, communication networks, and mechanical structures.

Properties of the Piano Wire

The given wire is characterized by its mass (m), length (L), and tension (T). The mass per unit length, μ, is a key parameter influencing wave speed and energy transmission. It is computed as:

μ = m / L

Given m = 3.00 g = 0.003 kg and L = 80.0 cm = 0.80 m, we find:

μ = 0.003 kg / 0.80 m = 0.00375 kg/m

Wave Properties

The wave has a frequency (f) of 120.0 Hz and an amplitude (A) of 1.6 mm = 0.0016 m. The wave speed (v) on the string is given by the relation:

v = √(T / μ)

Substituting values, we find:

v = √(25.0 N / 0.00375 kg/m) ≈ √6666.67 ≈ 81.65 m/s

Calculating the Power Carried by the Wave

The average power transmitted by a sinusoidal wave on a string is given by:

P = (1/2) μ ω² A² v

where ω = 2πf is the angular frequency. Calculating ω:

ω = 2π × 120.0 Hz ≈ 2 × 3.1416 × 120 ≈ 753.98 rad/s

Now, substituting the knowns, the power P is:

P = (1/2) × 0.00375 kg/m × (753.98 rad/s)² × (0.0016 m)² × 81.65 m/s

Calculating step-by-step:

  • ω² ≈ 753.98² ≈ 567,006
  • A² = (0.0016)² = 2.56 × 10⁻⁶ m²

Therefore:

P ≈ 0.5 × 0.00375 × 567,006 × 2.56 × 10⁻⁶ × 81.65

Computing further:

  • 0.5 × 0.00375 ≈ 0.001875
  • 567,006 × 2.56 × 10⁻⁶ ≈ 1.452
  • 0.001875 × 1.452 ≈ 0.002722
  • Finally, 0.002722 × 81.65 ≈ 0.222

Thus, the average power carried by the wave is approximately:

P ≈ 0.222 Watts

Effect of Halving the Wave Amplitude

Since the power P depends on the square of the amplitude A, when the amplitude is halved (A/2), the new power P' becomes:

P' = (1/2) μ ω² (A/2)² v = (1/2) μ ω² A² (1/2)² v = P × (1/4)

Therefore, the average power decreases by a factor of four. If the original power was approximately 0.222 W, then the new power with half the amplitude is:

P' ≈ 0.222 W / 4 ≈ 0.0555 W

In conclusion, halving the wave amplitude results in the average power being reduced to a quarter of its original value, illustrating the quadratic dependence of power on amplitude.

Conclusion

This analysis demonstrates how fundamental physical parameters influence wave power transmission along a stretched wire. The calculated average power of approximately 0.222 Watts reflects the energy transfer capability of the wave under given conditions. The significant impact of amplitude changes on power underscores the importance of precise control over wave parameters in practical applications like musical instruments and mechanical systems. Understanding these relationships enables better design and optimization of systems involving wave mechanics.

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