Chapter 14, Problem 14.1DQ

An object is moving with SHM of amplitude A on the end of a spring. If the amplitude is doubled, what happens to the total distance the object travels in one period? What happens to the period? What happens to the maximum speed of the object? Discuss how these answers are related.

To determine

The distance travelled by the object in one period, effect on time and maximum speed of the object.

Explanation of Solution

Section 1:

To determine: The distance travelled by the object in one period after making twice of the amplitude.

Answer: The distance travelled by the object in one period after making twice of the amplitude is $4A$.

Given Info: The motion is SHM for a object and the amplitude is $A$.

The simple harmonic motion is a kind of motion when the retarding force is directly proportional to the displacement by Hook’s Law.

Formula to calculate displacement of a particle in simple harmonic motion is,

$x=A\cos \left(\omega t+\varphi \right)$`

• $x$ is the displacement.
• $A$ is the amplitude.
• $\omega$ is the angular velocity of the object.
• $t$ is the time period.
• $\varphi$ is the phase angle in SHM.

Amplitude is the maximum displacement travelled by the object about is mean position.

From above relation it is visible that the displacement of any object is directly proportional to the amplitude of the motion. The displacement here ranges from $A$ to $-A$.

The distance travelled is,

$\begin{array}{c}x=A-\left(-A\right)\\ =2A\end{array}$

When we double the amplitude the displacement is,

Substitute $2A$ for $A$ in above equation to find $x$.

$\begin{array}{c}x=2\left(2A\right)\\ =4A\end{array}$

Thus, by increasing the amplitude by $2$, the distance travelled by the object in one period is $4A$.

Section 2:

To determine: The effect on time period after making twice of the amplitude.

Answer: The effect on time period after making twice of the amplitude is zero so the time is same.

Given Info: The motion is SHM for a object and the amplitude is $A$.

The equation of time period in SHM is,

$T=2\pi \sqrt{\frac{m}{k}}$

• $m$ is the mass of the object
• $k$ is the spring constant

From above equation it is clear that the time function is independent of the amplitude so by increasing or decreasing the magnitude no changes will be seen in the time of travel.

Thus, the effect on time period after making twice of the amplitude is zero so the time is same.

Section 3:

To determine: The maximum speed of the object.

Answer: The maximum speed of the object is doubled.

Given Info: The motion is SHM for a object and the amplitude is $A$.

The equation of velocity in SHM is,

$v_{\text{m}}=A\omega$

• $v_{\text{m}}$ is the maximum velocity of the object

The above relation indicates that by increasing the amplitude the velocity of object will also increase.

Thus, the maximum speed of the object is doubled after multiplying the amplitude by two.

Conclusion:

Therefore, the distance travelled by the object in one period is $4A$ , time period remains the same and maximum speed of the object is double to the previous one.