A string that is under 50N of tension has a linear density of 5.0 g/m. A sinusoidal wave with amplitude 3.0 cm and wavelength 2.0 m travels along the wave. What is the maximum velocity of a particle on the string? 9.42 m/s 18.84 m/s 100 m/s 300 m/s
A string that is under 50N of tension has a linear density of 5.0 g/m. A sinusoidal wave with amplitude 3.0 cm and wavelength 2.0 m travels along the wave. What is the maximum velocity of a particle on the string? 9.42 m/s 18.84 m/s 100 m/s 300 m/s
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![**Question:**
A string that is under 50N of tension has a linear density of 5.0 g/m. A sinusoidal wave with amplitude 3.0 cm and wavelength 2.0 m travels along the wave. What is the maximum velocity of a particle on the string?
**Options:**
- 9.42 m/s
- 18.84 m/s
- 100 m/s
- 300 m/s
- Impossible to tell from the information given
**Explanation:**
This question requires an understanding of wave mechanics and the relationship between tension, linear density, wave speed, and particle velocity.
### Solution:
To find the wave speed \( v \) on a string, use the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the string (50 N) and \( \mu \) is the linear density (5.0 g/m or 0.005 kg/m).
First, let's calculate the wave speed:
\[ v = \sqrt{\frac{50 \text{ N}}{0.005 \text{ kg/m}}} \]
\[ v = \sqrt{10000 \text{ m}^2/\text{s}^2} \]
\[ v = 100 \text{ m/s} \]
Next, using the wave speed, we can find the frequency \( f \) since the wavelength \( \lambda \) is given (2.0 m):
\[ f = \frac{v}{\lambda} \]
\[ f = \frac{100 \text{ m/s}}{2.0 \text{ m}} \]
\[ f = 50 \text{ Hz} \]
The maximum velocity \( v_{\text{max}} \) of a particle on the string is given by:
\[ v_{\text{max}} = \omega A \]
where \( \omega \) is the angular frequency (\( \omega = 2 \pi f \)) and \( A \) is the amplitude (3.0 cm or 0.03 m).
First, calculate the angular frequency \( \omega \):
\[ \omega = 2 \pi \times 50 \text{ Hz} \]
\[ \omega = 100 \pi \text{ rad/s} \]
Then, calculate the maximum velocity:
\[ v_{\text{max}} = 100 \pi \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdfe80350-6fa1-469d-aae2-95b4dd3aa499%2Ff2eb0715-de49-40fc-96a4-ae1bef7303e6%2Fim4cn3_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
A string that is under 50N of tension has a linear density of 5.0 g/m. A sinusoidal wave with amplitude 3.0 cm and wavelength 2.0 m travels along the wave. What is the maximum velocity of a particle on the string?
**Options:**
- 9.42 m/s
- 18.84 m/s
- 100 m/s
- 300 m/s
- Impossible to tell from the information given
**Explanation:**
This question requires an understanding of wave mechanics and the relationship between tension, linear density, wave speed, and particle velocity.
### Solution:
To find the wave speed \( v \) on a string, use the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the string (50 N) and \( \mu \) is the linear density (5.0 g/m or 0.005 kg/m).
First, let's calculate the wave speed:
\[ v = \sqrt{\frac{50 \text{ N}}{0.005 \text{ kg/m}}} \]
\[ v = \sqrt{10000 \text{ m}^2/\text{s}^2} \]
\[ v = 100 \text{ m/s} \]
Next, using the wave speed, we can find the frequency \( f \) since the wavelength \( \lambda \) is given (2.0 m):
\[ f = \frac{v}{\lambda} \]
\[ f = \frac{100 \text{ m/s}}{2.0 \text{ m}} \]
\[ f = 50 \text{ Hz} \]
The maximum velocity \( v_{\text{max}} \) of a particle on the string is given by:
\[ v_{\text{max}} = \omega A \]
where \( \omega \) is the angular frequency (\( \omega = 2 \pi f \)) and \( A \) is the amplitude (3.0 cm or 0.03 m).
First, calculate the angular frequency \( \omega \):
\[ \omega = 2 \pi \times 50 \text{ Hz} \]
\[ \omega = 100 \pi \text{ rad/s} \]
Then, calculate the maximum velocity:
\[ v_{\text{max}} = 100 \pi \
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