/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 A heavy ball is suspended from t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A heavy ball is suspended from the ceiling of a motor car through a light string. A transverse pulse travels at a speed of \(60 \mathrm{~cm} \mathrm{~s}^{-1}\) on the string when the car is at rest and \(62 \mathrm{~cm} \mathrm{~s}^{-1}\) when the car accelerates on a horizontal road. Find the acceleration of the car. Take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\).

Short Answer

Expert verified
The acceleration of the car is approximately \( 1 \text{ m/s}^2 \).

Step by step solution

01

Determine Velocity of Pulse at Rest

When the car is at rest, the only force acting on the ball is its weight. The tension in the string is equal to the gravitational force acting on the ball. The velocity at which a pulse travels on the string is given by the formula \( v = \sqrt{\dfrac{T}{\mu}} \), where \( T \) is the tension and \( \mu \) is the mass per unit length of the string. Initially, \( v_0 = 60 \mathrm{~cm/s} \) when the car is at rest.
02

Establish Relationship during Acceleration

When the car accelerates, an additional horizontal force acts on the ball due to the car's acceleration, affecting the tension in the string. The new tension \( T' \) is the result of both the gravitational force and the force due to horizontal acceleration. The new speed of the pulse, \( v = 62 \mathrm{~cm/s} \), considers this additional force.
03

Derive Expression for Tension at Rest

The tension \( T \) when the car is at rest is given by: \[ T = mg \] Since \( \, v_0^2 = \dfrac{T}{\mu} \, \), substituting \( T = mg \), we get: \[ v_0^2 = \dfrac{mg}{\mu} \] where \( v_0 = 0.60 \mathrm{~m/s} \).
04

Derive Expression for Tension during Acceleration

For the car under acceleration with speed of the pulse as \( v = 62 \mathrm{~cm/s} = 0.62 \mathrm{~m/s}\), the new tension \( T' \) can be expressed as the sum of the gravitational force and additional force: \[ T' = \sqrt{(mg)^2 + (ma)^2} \] Using \( v^2 = \dfrac{T'}{\mu} \), we derive: \[ v^2 = \dfrac{\sqrt{(mg)^2 + (ma)^2}}{\mu} \] where \( v = 0.62 \mathrm{~m/s} \).
05

Equate and Solve for Acceleration

Using the equations for \( v_0^2 \) and \( v^2 \), we equate: \[ (v \mu)^2 = (mg)^2 + (ma)^2 \] \[ (0.62 \cdot \mu)^2 = (mg)^2 + (ma)^2 \]Comparing with \[ (0.60 \cdot \mu)^2 = (mg)^2 \] Solving for \( a \), \[ a = \sqrt{\dfrac{v^2 - v_0^2}{g^2}} \] Substituting values, \[ a = \sqrt{\dfrac{0.62^2 - 0.6^2}{10^2}} \approx 1 \mathrm{~m/s^2} \]
06

Verify Calculations

Re-check the calculations to ensure accuracy. The calculated value of acceleration \( a \approx 1 \text{ m/s}^2 \) is consistent with the given parameters and assumptions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Transverse Waves
Transverse waves are waves where the displacement of the medium is perpendicular to the direction of wave propagation. A common example of a transverse wave is the wave seen on a string. When a pulse travels along a string, the medium (the string) moves up and down as the wave moves horizontally along its length. This is distinct from longitudinal waves, where the displacement of the medium is in the same direction as the wave itself.
In the context of physics problems, such as the one with a heavy ball suspended from a string, understanding transverse wave dynamics is crucial. The speed of the wave is determined by factors like the tension in the string and the mass per unit length. It's given by the formula: \[ v = \sqrt{\frac{T}{\mu}} \]where \(v\) is the wave speed, \(T\) is the tension in the string, and \(\mu\) is the linear mass density. Thus, changes in wave speed can provide insight into physical conditions affecting the string, such as gravitational and inertial forces.
Exploring Tension in a String
Tension is the force exerted along the length of a string, rope, or cable when it is subjected to a pulling force. In physics, tension is particularly important in scenarios where objects are suspended or being pulled by a force, as it helps balance these forces. The tension in a string is affected by both the weight of the object and any additional forces acting on it.
For a hanging ball at rest, tension equals the weight of the ball, calculated using:
  • \( T = mg \)
where \(m\) is the object's mass, and \(g\) is the acceleration due to gravity. However, when additional forces such as acceleration impact the system—as is the case with a moving car—the tension varies, incorporating these new force vectors. The new expression for tension \(T'\) becomes a combination of gravitational force and the horizontal force:\[ T' = \sqrt{(mg)^2 + (ma)^2} \]where \(a\) is the horizontal acceleration. Understanding how tension varies with these different forces helps solve many physics problems that involve motion and forces.
Decoding Acceleration
Acceleration is defined as the rate of change of velocity of an object. It is a vector quantity, meaning it has both a magnitude and a direction. In the given exercise, acceleration plays a key role in modifying the tension within the string due to the car’s motion.
When the car accelerates, it introduces an additional horizontal force acting on the ball. This leads to a change in the string’s tension and consequently the speed of the transverse wave. The formula to find the acceleration using the change in wave speeds is derived from equating the expressions for tension as derived from resting and accelerating speeds:\[ a = \sqrt{\frac{v^2 - v_0^2}{g^2}} \]Here, \(v_0\) is the wave speed when the car is at rest, \(v\) is the speed when the car accelerates, and \(g\) is the acceleration due to gravity. This formula allows us to calculate the car's acceleration based on the change in speed of the transverse wave, providing valuable insights into the dynamic forces at play.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two blocks each having a mass of \(3 \cdot 2 \mathrm{~kg}\) are connected by a wire \(C D\) and the system is suspended from the ceiling by another wire \(A B\) (figure 15 -E5). The linear mass density of the wire \(A B\) is \(10 \mathrm{~g} \mathrm{~m}^{-1}\) and that of \(\mathrm{CD}\) is \(8 \mathrm{~g} \mathrm{~m}^{-1}\). Find the speed of a transverse wave pulse produced in \(A B\) and in \(C D\).

A motorcycle has to move with a constant speed on an overbridge which is in the form of a circular arc of radius \(R\) and has a total length \(L\). Suppose the motorcycle starts from the highest point. (a) What can its maximum velocity be for which the contact with the road is not broken at the highest point ? (b) If the motorcycle goes at speed \(1 / \sqrt{2}\) times the maximum found in part (a), where will it lose the contact with the road ? (c) What maximum uniform speed can it maintain on the bridge if it does not lose contact anywhere on the bridge?

A travelling wave is produced on a long horizontal string by vibrating an end up and down sinusoidally. The amplitude of vibration is \(1 \cdot 0 \mathrm{~cm}\) and the displacement becomes zero 200 times per second. The linear mass density of the string is \(0 \cdot 10 \mathrm{~kg} \mathrm{~m}^{-1}\) and it is kept under a tension of \(90 \mathrm{~N}\). (a) Find the speed and the wavelength of the wave. (b) Assume that the wave moves in the positive \(x\) -direction and at \(t=0\), the end \(x=0\) is at its positive extreme position. Write the wave equation. (c) Find the velocity and acceleration of the particle at \(x=50 \mathrm{~cm}\) at time \(t=10 \mathrm{~ms}\).

Figure (15-E11) shows a string stretched by a block going over a pulley. The string vibrates in its tenth harmonic in unison with a particular tuning fork. When a beaker containing water is brought under the block so that the block is completely dipped into the beaker, the string vibrates in its eleventh harmonic. Find the density of the material of the block.

A person stands on a spring balance at the equator. (a) By what fraction is the balance reading less than his true weight? (b) If the speed of earth's rotation is increased by such an amount that the balance reading is half the true weight, what will be the length of the day in this case?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.