Problem 37 - College Physics

Question

\(\bullet\bullet\) The equation of motion of a particle in vertical SHM is given by \(y=(10 \mathrm{~cm}) \sin [(0.50 \mathrm{rad} / \mathrm{s}) t]\). What are the particle's (a) displacement (b) velocity, and (c) acceleration at \(t=1.0 \mathrm{~s}\) ?

Step-by-Step Solution

Answer

Displacement: \( 4.79 \) cm, Velocity: \( 4.39 \) cm/s, Acceleration: \( -1.20 \) cm/s² at \( t=1.0 \) s.

Step 1: Identify Displacement Function

The displacement function for the particle is given by \( y(t) = 10 \sin(0.50t) \), where \( y \) is in centimeters and \( t \) is in seconds.

Step 2: Calculate Displacement at t=1.0s

To find the displacement at \( t=1.0 \) s, substitute \( t=1.0 \) into the displacement equation: \( y(1) = 10 \sin(0.50 \times 1) = 10 \sin(0.50) \). Calculate \( \sin(0.50) \), then multiply by 10 cm.

Step 3: Derive the Velocity Function

The velocity \( v(t) \) is the derivative of the displacement function \( y(t) \). Differentiate \( y(t) = 10 \sin(0.50t) \) with respect to \( t \): \( v(t) = 10 \times 0.50 \cos(0.50t) = 5 \cos(0.50t) \).

Step 4: Calculate Velocity at t=1.0s

To find the velocity at \( t=1.0 \) s, substitute \( t=1.0 \) into the velocity equation: \( v(1) = 5 \cos(0.50 \times 1) = 5 \cos(0.50) \). Calculate \( \cos(0.50) \), then multiply by 5 cm/s.

Step 5: Derive the Acceleration Function

The acceleration \( a(t) \) is the derivative of the velocity function \( v(t) \). Differentiate \( v(t) = 5 \cos(0.50t) \) with respect to \( t \): \( a(t) = -5 \times 0.50 \sin(0.50t) = -2.5 \sin(0.50t) \).

Step 6: Calculate Acceleration at t=1.0s

To find the acceleration at \( t=1.0 \) s, substitute \( t=1.0 \) into the acceleration equation: \( a(1) = -2.5 \sin(0.50 \times 1) = -2.5 \sin(0.50) \). Calculate \( \sin(0.50) \) again, then multiply by -2.5 cm/s².

Key Concepts

Displacement in Simple Harmonic Motion

In the context of Simple Harmonic Motion (SHM), displacement refers to how far a particle is from its mean or equilibrium position at a given time. In our exercise, the particle's displacement is described using the equation \( y(t) = 10 \sin(0.50t) \), where \( y \) is the displacement in centimeters, and \( t \) is time in seconds. This equation shows that the particle moves in a sinusoidal wave pattern—a characteristic of SHM—where the amplitude, the maximum displacement from the equilibrium, is 10 cm. To calculate the displacement at a specific time, say \( t=1.0 \) second, you substitute \( t \) into the equation. For this exercise:

• Substitute \( t=1.0 \) into the displacement equation to get \( y(1) = 10 \sin(0.50) \).
• You will need to calculate \( \sin(0.50) \) and then multiply it by 10 cm to find the displacement at that time.

This step shows that the position of the particle is periodic and continuously varies with time, making it predictable given the function.

Velocity in Simple Harmonic Motion

Velocity is the rate at which the particle’s position changes over time. In SHM, this is determined by differentiating the displacement function. For our particle, the velocity equation derived from \( y(t) = 10 \sin(0.50t) \) becomes \( v(t) = 5 \cos(0.50t) \) following the differentiation rules. It is key to understand that the particle’s velocity also oscillates similarly to its displacement over time.To find the velocity at \( t=1.0 \) second, use the following steps:

• Substitute \( t=1.0 \) into the velocity function: \( v(1) = 5 \cos(0.50) \).
• Determine \( \cos(0.50) \) and then multiply by 5 cm/s to get the velocity at \( t=1.0 \) second.

In SHM, velocity reaches its maximum when the particle passes through its equilibrium position. Conversely, it is zero at the points of maximum displacement, indicating a change in direction.

Acceleration in Simple Harmonic Motion

Acceleration describes how quickly the velocity of the particle changes. In SHM, the acceleration is directly related to the displacement but in the opposite direction. It can be found by differentiating the velocity function. From the given velocity \( v(t) = 5 \cos(0.50t) \), the acceleration function is derived as \( a(t) = -2.5 \sin(0.50t) \).To calculate the acceleration at \( t=1.0 \) second:

• Substitute \( t=1.0 \) into the acceleration equation: \( a(1) = -2.5 \sin(0.50) \).
• Calculate \( \sin(0.50) \) and then multiply by \(-2.5\) cm/s² to find the particle's acceleration at one second.

Acceleration in SHM is always directed towards the mean position, emphasizing the particle's tendency to return to equilibrium. This restoring force is what gives SHM its characteristic repetitive motion.