# Given Data - Wave speed, \( v = 10.\ \mathrm{cm/s} \) - Wavelength, \( \lambda = 25.\ \mathrm{cm} \) - At \( t = \), \( x = 5.00\ \mathrm{cm} \), displacement \( y = -80.\ \mathrm{cm} \) (maximum downward) - Need to find: - (a) Equations for position \( y(x, t) \) and velocity \( v(x, t) \) - (b) Position and velocity at \( x = 5\ \mathrm{cm},\ t = .25\ \mathrm{s} \) --- ## Step 1: Write the Position and Velocity Equations ### **A. Position Equation** The general form for a sinusoidal transverse wave is: \[ y(x, t) = A \sin(kx - \omega t + \phi) \] where: - \( A \) = amplitude - \( k = \frac{2\pi}{\lambda} \) = wavenumber - \( \omega = 2\pi f \) = angular frequency - \( v = f\lambda \implies f = \frac{v}{\lambda} \) Let's calculate each term: - \( k = \frac{2\pi}{25.} = .251\ \mathrm{rad/cm} \) - \( f = \frac{10.}{25.} = .4\ \mathrm{Hz} \) - \( \omega = 2\pi \times .4 = .8\pi \approx 2.513\ \mathrm{rad/s} \) Given at \( t=, x=5.\,\mathrm{cm}, y=-80.\,\mathrm{cm} \) (maximum downward), so \( A = 80.\,\mathrm{cm} \). Maximum displacement occurs at \(\sin(\theta) = -1\): \[ -80. = 80.\sin(k \cdot 5. + \phi) \implies \sin(k \cdot 5. + \phi) = -1 \implies k \cdot 5. + \phi = \frac{3\pi}{2} \] So, \[ \phi = \frac{3\pi}{2} - k \cdot 5. \] Plug in \(k\): \[ k \cdot 5. = .251 \times 5 = 1.257\ \mathrm{rad} \] \[ \phi = \frac{3\pi}{2} - 1.257 \approx 4.712 - 1.257 = 3.455\ \mathrm{rad} \] **Position Equation:** \[ \boxed{ y(x, t) = 80.\,\sin\left[.251\,x - 2.513\,t + 3.455\right]\ \mathrm{cm} } \] **Explanation:** We used the general equation for a transverse wave, calculated wavenumber and angular frequency from given values, and used the initial condition to find the phase constant. --- ### **B. Velocity Equation** The transverse velocity is the time derivative: \[ v(x, t) = \frac{\partial y}{\partial t} = -A\omega \cos(kx - \omega t + \phi) \] Plug in values: \[ v(x, t) = -80. \times 2.513 \cos(.251\,x - 2.513\,t + 3.455) \] \[ v(x, t) = -201.\,\cos(.251\,x - 2.513\,t + 3.455)\ \mathrm{cm/s} \] **Explanation:** Time-derivative of position gives velocity. We substitute the known amplitude and angular frequency. --- ## Step 2: Find Position and Velocity at \( x 5\,\mathrm{cm},\ t = .25\,\mathrm{s} \) ### **A. Position at \( x=5, t=.25 \)** Plug into position equation: \[ y(5, .25) = 80.\,\sin\left[.251 \times 5 - 2.513 \times .25 + 3.455\right] \] Calculate inside the brackets: - \( .251 \times 5 = 1.257 \) - \( 2.513 \times .25 = .628 \) - So, argument \( = 1.257 - .628 + 3.455 = 4.084 \ \mathrm{rad} \) Now, \[ y(5, .25) = 80. \times \sin(4.084) \] \( \sin(4.084) \approx -.818 \) \[ y(5, .25) = 80. \times (-.818) \approx -65.4\ \mathrm{cm} \] **Explanation:** We substitute \( x \) and \( t \) into the position equation and evaluate. --- ### **B. Velocity at \( x=5, t=.25 \)** Plug into velocity equation: \[ v(5, .25) = -201.\,\cos(4.084) \] \( \cos(4.084) \approx -.574 \) \[ v(5, .25) = -201. \times (-.574) \approx 115.3\ \mathrm{cm/s} \] **Explanation:** Substitute same values into the velocity equation and evaluate cosine. --- ## Final Answers ### (a) **Position and Velocity Equations** \[ y(x, t) = 80.\,\sin\left(.251\,x - 2.513\,t + 3.455\right)\ \mathrm{cm} \] \[ v(x, t) = -201.\,\cos\left(.251\,x - 2.513\,t + 3.455\right)\ \mathrm{cm/s} \] --- ### (b) **At \( x = 5\,\mathrm{cm},\ t = .25\,\mathrm{s} \):** - **Position:** \( y = -65.4\,\mathrm{cm} \) - **Velocity:** \( v = 115.3\,\mathrm{cm/s} \)