Question

A string clamped at both the ends has a mass 10gm. length 1m and it is kept under tension of 1N. It is vibrating in the fundamental mode with an amplitude of 1cm .Assuming the standing wave pattern, the maximum acceleration seen in the string is

• A. 4\(\pi\)² m/s²
• B. 2\(\pi\)² m/s²
• C. \(\pi\)² m/s²
• D. 4\(\pi\) m/s²
• E. 2\(\pi\) m/s²

Answer 1

The Correct Option is C

The fundamental frequency \( f_1 \) of a string clamped at both ends is given by:
\[ f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

where:
- \( L \) is the length of the string (1 m),
- \( T \) is the tension in the string (1 N),
- \( \mu \) is the linear mass density of the string (\(\frac{\text{mass}}{\text{length}} = \frac{10 \text{ gm}}{1 \text{ m}} = 0.01 \text{ kg/m}\)).

Let's calculate the fundamental frequency:
\[ f_1 = \frac{1}{2 \times 1} \sqrt{\frac{1}{0.01}} = \frac{1}{2} \sqrt{100} = \frac{1}{2} \times 10 = 5 \text{ Hz} \]

The angular frequency \( \omega \) is:
\[ \omega = 2 \pi f_1 = 2 \pi \times 5 = 10 \pi \text{ rad/s} \]

The displacement of the string in the fundamental mode can be described as a standing wave with nodes at the ends and an antinode in the middle. The displacement \( y(x,t) \) is given by:
\[ y(x,t) = A \sin \left( \frac{\pi x}{L} \right) \cos (\omega t) \]

where \( A \) is the amplitude of the wave (1 cm = 0.01 m).

The acceleration \( a \) is the second time derivative of the displacement:
\[ a(x,t) = \frac{\partial^2 y(x,t)}{\partial t^2} \]
\[ a(x,t) = -A \sin \left( \frac{\pi x}{L} \right) \omega^2 \cos (\omega t) \]

The maximum acceleration occurs at the antinode (middle of the string, \( x = \frac{L}{2} \)) where \( \sin \left( \frac{\pi x}{L} \right) = 1 \) and \( \cos (\omega t) = \pm 1 \):

\[ a_{\text{max}} = A \omega^2 \]
\[ a_{\text{max}} = 0.01 \times (10 \pi)^2 \]
\[ a_{\text{max}} = 0.01 \times 100 \pi^2 \]
\[ a_{\text{max}} = \pi^2 \text{ m/s}^2 \]
Thus the correct Answer is Option C \( \pi^2 \text{ m/s}^2\). 

Thus, the maximum acceleration in the string should be \( \pi^2 \text{ m/s}^2\).

Answer 2

1. Find the fundamental frequency (f):

For a string clamped at both ends, the fundamental frequency is given by:

\[f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}\]

where:

• L = 1 m (length of the string)
• T = 1 N (tension)
• μ = m/L = (0.01 kg) / (1 m) = 0.01 kg/m (linear mass density)

\[f = \frac{1}{2(1)}\sqrt{\frac{1}{0.01}} = \frac{1}{2}\sqrt{100} = \frac{10}{2} = 5 \, Hz\]

2. Find the angular frequency (ω):

\[\omega = 2\pi f = 2\pi(5) = 10\pi \, \text{rad/s}\]

3. Find the maximum acceleration:

The equation for the displacement of a point on the string vibrating in the fundamental mode is given by:

\[y(x,t) = A\sin(kx)\cos(\omega t)\]

where A is the amplitude. The acceleration is the second derivative of displacement with respect to time:

\[a(x,t) = \frac{\partial^2 y}{\partial t^2} = -\omega^2 A\sin(kx)\cos(\omega t)\]

The maximum acceleration occurs when \(\sin(kx) = 1\) and \(\cos(\omega t) = 1\):

\[a_{max} = \omega^2 A = (10\pi)^2 (0.01 \, m) = 100\pi^2 (0.01) = \pi^2 \, m/s^2\]