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# test-1 - far away is the cave wall Use for the speed of...

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Test # 1 Solutions 1. A person is creating a standing wave in a cylinder that is open at one end and closed at the other end. The height of the cylinder is 50 cm. What is the wavelength of the fundamental standing wave? Given the speed of sound, 340 m/s, what is the frequency of this wave? Sol 1. With this configuration, we can fit a quarter of a wavelength of the funda- mental standing wave. Therefore, the wavelength of the fundamental standing wave is simply λ = 4 × (0 . 5m) = 2 . 0m It is then straightforward to compute the frequency of this wave f = v/λ = (340m/s) / (2 . 0m) = 170s - 1 2. A skipper on a boat notices wave crests passing his anchor chain every 4 s. He estimates the distance between the wave crests to be 16 m. What is the speed of the waves based on his estimates? Sol 2. Based on these estimates, the speed of the waves is v = (distance between wave crests) / (time between two wave crests) = (16m) / (4s) = 4m/s 3. A bat flying in a cave emits a sound and receives its echo 0.2 s later. How
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Unformatted text preview: far away is the cave wall? Use for the speed of sound, 340 m/s. Sol 3. The emitted sound wave travels twice the distance between the bat and the wall, in hitting the wall and echoing back to the bat. We assume that the speed at which the bat is ﬂying is much smaller than the speed of sound, so that the position of the bat has not varied appreciably with respect to the wall during these 0.2 s. Under this assumption, the distance to the cave wall is given by d = 1 2 (340m/s × . 2s) = 34m 4. By what distance must two charges of +4 C be separated so that the repulsive force between them is 3 . 6 × 10 10 N. (Coulomb’s constant, k = 9 × 10 9 Nm 2 / C 2 .) Sol 4. We use the Coulomb force law F = k q 1 q 2 r 2 ⇒ r = √ kq 1 q 2 F = ± (9 × 10 9 Nm 2 C-2 ) · (+4C) · (+4C) 3 . 6 × 10 10 N = 2m 1...
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