
A small piece of cork in a ripple tank oscillates up and down as ripples pass it. If the ripples travelling at 0.3 m/s have a wavelength of \[1.5\pi \] cm and the cork vibrates with an amplitude of 5 mm, the maximum velocity of the cork is:
A. 20 cm/sec
B. 20 m/sec
C. 0.02 m/sec
D. 200 cm/sec
Answer
162.3k+ views
Hint:The cork will oscillate along the vertical direction as the ripple travels. For the cork we use sine wave function to find the velocity or displacement. The maximum velocity of the cork will be at the mean equilibrium position.
Formula used:
\[v = \nu \lambda \]
where v is the wave speed, \[\nu \] is the frequency of the wave and \[\lambda \] is the wavelength of the wave.
\[{v_{\max }} = \omega A\]
where \[{v_{\max }}\] is the maximum speed of simple harmonic motion, \[\omega \] is the angular frequency and A is the amplitude of the oscillation.
Complete step by step solution:
Given the velocity of the ripples is 0.3 m/s
\[v = 0.3m/s\]
Wavelength of the ripple is \[1.5\pi \] cm
\[\lambda = 1.5\pi \times {10^{ - 2}}m\]
Amplitude of the displacement of the cork is 5 mm. That means the cork will have a maximum vertical displacement of 5 mm from the mean equilibrium position.
\[A = 5 \times {10^{ - 3}}m\]
Using the relation \[v = \nu \lambda \], the frequency of the oscillation will be,
\[\nu = \dfrac{v}{\lambda }\]
Putting the values, we get
\[\nu = \dfrac{{0.3}}{{1.5\pi \times {{10}^{ - 2}}}}Hz \\ \]
\[\Rightarrow \nu = \dfrac{{20}}{\pi }Hz\]
Then the angular frequency of the oscillation will be,
\[\omega = 2\pi \nu \]
\[\Rightarrow \omega = 2\pi \times \dfrac{{20}}{\pi }\] rad/s
\[\Rightarrow \omega = 40\] rad/s
The maximum wave speed is equal to \[{v_{\max }} = \omega A\]
Putting the values, we get
\[{v_{\max }} = 40 \times 5 \times {10^{ - 3}}\] m/s
\[\Rightarrow {v_{\max }} = 0.2\] m/s
\[\Rightarrow {v_{\max }} = 0.2 \times 100\] cm/s
\[\therefore {v_{\max }} = 20\] cm/s
Hence, the maximum speed of the cork is 20 cm/s.
Therefore, the correct option is A.
Note: The ripple travels along the radial direction and the cork is oscillating in the vertical plane. The speed of the ripple is constant but the speed of the cork varies as the displacement from the mean equilibrium position.
Formula used:
\[v = \nu \lambda \]
where v is the wave speed, \[\nu \] is the frequency of the wave and \[\lambda \] is the wavelength of the wave.
\[{v_{\max }} = \omega A\]
where \[{v_{\max }}\] is the maximum speed of simple harmonic motion, \[\omega \] is the angular frequency and A is the amplitude of the oscillation.
Complete step by step solution:
Given the velocity of the ripples is 0.3 m/s
\[v = 0.3m/s\]
Wavelength of the ripple is \[1.5\pi \] cm
\[\lambda = 1.5\pi \times {10^{ - 2}}m\]
Amplitude of the displacement of the cork is 5 mm. That means the cork will have a maximum vertical displacement of 5 mm from the mean equilibrium position.
\[A = 5 \times {10^{ - 3}}m\]
Using the relation \[v = \nu \lambda \], the frequency of the oscillation will be,
\[\nu = \dfrac{v}{\lambda }\]
Putting the values, we get
\[\nu = \dfrac{{0.3}}{{1.5\pi \times {{10}^{ - 2}}}}Hz \\ \]
\[\Rightarrow \nu = \dfrac{{20}}{\pi }Hz\]
Then the angular frequency of the oscillation will be,
\[\omega = 2\pi \nu \]
\[\Rightarrow \omega = 2\pi \times \dfrac{{20}}{\pi }\] rad/s
\[\Rightarrow \omega = 40\] rad/s
The maximum wave speed is equal to \[{v_{\max }} = \omega A\]
Putting the values, we get
\[{v_{\max }} = 40 \times 5 \times {10^{ - 3}}\] m/s
\[\Rightarrow {v_{\max }} = 0.2\] m/s
\[\Rightarrow {v_{\max }} = 0.2 \times 100\] cm/s
\[\therefore {v_{\max }} = 20\] cm/s
Hence, the maximum speed of the cork is 20 cm/s.
Therefore, the correct option is A.
Note: The ripple travels along the radial direction and the cork is oscillating in the vertical plane. The speed of the ripple is constant but the speed of the cork varies as the displacement from the mean equilibrium position.
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