A tuning fork with a frequency of 440 Hz is connected to a long string having a linear mass density of 0.01 kg m^ -1 and maintained under a tension of 49 N . The tuning fork generates transverse waves with an amplitude

Correct answer: A. 0.67 W Given: Frequency f = 440 Hz Linear mass density μ = 0.01 kg m^ -1 Tension T = 49 N Amplitude A = 0.50 mm = 0.5 × 10^ -3 m The wave speed v is given by: v = (T)/(μ) = (49)/(0.01) = 70 m s^ -1 The angular frequency ω is: ω = 2 π f = 2 π × 440 = 880 π rad s^ -1 The average rate at which energy is transmitted (average power) is: P_ avg = (1)/(2) μ v ω^2 A^2 Substituting the values: P_ avg = (1)/(2) × 0.01 × 70 × (880 π)^2 × (0.5 × 10^ -3 )^2 P_ avg = 0.35 × 774400 π^2 × 0.25 × 10^ -6 P_ avg = 0.06776 π^2 Using π^2 9.87 : P_ avg 0.06776 × 9.87 0.67 W

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Wave Motion and Waves on a String

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Wave Motion and Waves on a String

A tuning fork with a frequency of 440 Hz is connected to a long...

A tuning fork with a frequency of

440 \text{ Hz}

is connected to a long string having a linear mass density of

0.01 \text{ kg m}^{-1}

and maintained under a tension of

49 \text{ N}

. The tuning fork generates transverse waves with an amplitude of

0.50 \text{ mm}

along the string. Determine the average rate at which the tuning fork transmits energy to the string.

Options

A $0.67 \text{ W}$
B $2.68 \text{ W}$
C $1.34 \text{ W}$
D $0.34 \text{ W}$

Correct answer

A. $0.67 \text{ W}$

Step-by-step solution

Given:
Frequency

f = 440 \text{ Hz}

Linear mass density

\mu = 0.01 \text{ kg m}^{-1}

Tension

T = 49 \text{ N}

Amplitude

A = 0.50 \text{ mm} = 0.5 \times 10^{-3} \text{ m}

The wave speed

v

is given by:

v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{49}{0.01}} = 70 \text{ m s}^{-1}

The angular frequency

\omega

is:

\omega = 2 \pi f = 2 \pi \times 440 = 880 \pi \text{ rad s}^{-1}

The average rate at which energy is transmitted (average power) is:

P_{avg} = \dfrac{1}{2} \mu v \omega^2 A^2

Substituting the values:

P_{avg} = \dfrac{1}{2} \times 0.01 \times 70 \times (880 \pi)^2 \times (0.5 \times 10^{-3})^2

P_{avg} = 0.35 \times 774400 \pi^2 \times 0.25 \times 10^{-6}

P_{avg} = 0.06776 \pi^2

Using

\pi^2 \approx 9.87

P_{avg} \approx 0.06776 \times 9.87 \approx 0.67 \text{ W}

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