The displacement y of a particle executing simple harmonic motion is given by the equation:

The displacement y of a particle executing simple harmonic motion is given by the equation: y = 4 (0.2t + 0.3) This equation is in the standard form y = A ( t + ), where: • A is the amplitude • is the angular frequency • is the phase constant From the given equation, we can identify: Amplitude, A = 4 m Angular frequency, = 0.2 rad/s Phase constant, = 0.3 rad (i) Calculate the angular frequency, the frequency, period and phase of the motion. Step 1: Identify the angular frequency. From the equation, the angular frequency is directly given. = 0.2 rad/s The angular frequency is 0.2 rad/s. Step 2: Calculate the frequency (f). The frequency is related to the angular frequency by the formula f = ()/(2). f = 0.2 rad/s2 rad f = (0.1)/() Hz f ≈ (0.1)/(3.14159) Hz f ≈ 0.0318 Hz The frequency is 0.0318 Hz. Step 3: Calculate the period (T). The period is the reciprocal of the frequency, T = (1)/(f), or can be calculated as T = (2)/(). T = 2 rad0.2 rad/s T = 10 s T ≈ 10 × 3.14159 s T ≈ 31.4 s The period is 31.4 s. Step 4: Identify the phase of the motion. The phase constant (or initial phase) is directly given in the equation. = 0.3 rad The phase is 0.3 rad. (ii) Calculate the initial velocity and acceleration of the particle. Step 1: Find the general expression for velocity (v). Velocity is the first derivative of displacement with respect to time. v = (dy)/(dt) = (d)/(dt) [A ( t + )] v = A ( t + ) Substitute the values: v = (4 m)(0.2 rad/s) (0.2t + 0.3) v = 0.8 (0.2t + 0.3) m/s Step 2: Calculate the initial velocity (v_0) at t=0. v_0 = 0.8 (0.2(0) + 0.3) v_0 = 0.8 (0.3) m/s Using (0.3 rad) ≈ 0.9553: v_0 ≈ 0.8 × 0.9553 m/s v_0 ≈ 0.764 m/s The initial velocity is 0.764 m/s. Step 3: Find the general expression for acceleration (a). Acceleration is the first derivative of velocity with respect to time. a = (dv)/(dt) = (d)/(dt) [A ( t + )] a = -A^2 ( t + ) Substitute the values: a = -(4 m)(0.2 rad/s)^2 (0.2t + 0.3) a = -4 × 0.04 (0.2t + 0.3) m/s^2 a = -0.16 (0.2t + 0.3) m/s^2 Step 4: Calculate the initial acceleration (a_0) at t=0. a_0 = -0.16 (0.2(0) + 0.3) a_0 = -0.16 (0.3) m/s^2 Using (0.3 rad) ≈ 0.2955: a_0 ≈ -0.16 × 0.2955 m/s^2 a_0 ≈ -0.0473 m/s^2 The initial acceleration is -0.0473 m/s^2. (iii) Calculate the maximum velocity and acceleration. Step 1: Calculate the maximum velocity (v_max). The maximum velocity occurs when ( t + ) = ± 1. v_max = A v_max = (4 m)(0.2 rad/s) v_max = 0.8 m/s The maximum velocity is 0.8 m/s. Step 2: Calculate the maximum acceleration (a_max). The maximum acceleration occurs when ( t + ) = ± 1. a_max = A^2 a_max = (4 m)(0.2 rad/s)^2 a_max = 4 × 0.04 m/s^2 a_max = 0.16 m/s^2 The maximum acceleration is 0.16 m/s^2.