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

Question

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

Asked on March 29, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-03-29T09:50:39.300Z

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

Accepted Answer · 2026-03-29T09:50:39.300Z

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