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romanna [79]
3 years ago
11

Two trains on separate tracks move toward each other. Train 1 has a speed of 125 km/h; train 2, a speed of 94.0 km/h. Train 2 bl

ows its horn, emitting a frequency of 500 Hz. What is the frequency heard by the engineer on train 1?
Physics
1 answer:
dexar [7]3 years ago
4 0

Answer:

f_{o}=595.98Hz

Explanation:

Given data

Train 1 speed v₀=125 km/h =34.7222 m/s

Train 2 speed vs=94.0 km/h=26.1111 m/s

Speed of sound v=343 m/s

Train 2 frequency fs=500 Hz

To find

Frequency f₀ heard by engineer on train 1

Solution

We can use below formula to find the required frequency

f_{o}=f_{s}(\frac{v+v_{o}}{v-v_{s}} )\\f_{o}=(500Hz)(\frac{343m/s+34.722m/s}{343m/s-26.111m/s} ) \\f_{o}=595.98Hz

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Answer:

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Explanation:

Let m and a denote the mass and acceleration of Spiderman, respectively.

There are two forces on Spiderman:

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The directions of these two forces are exactly opposite of one another. Besides, because Spiderman is accelerating upwards, the magnitude of F(\text{tension}) (which points upwards) should be greater than that of W (which points downwards towards the ground.)

Subtract the smaller force from the larger one to find the net force on Spiderman:

(\text{Net Force}) = F(\text{tension}) - W.

On the other hand, apply Newton's Second Law of motion to find the value of the net force on Spiderman:

(\text{Net Force}) = m \cdot a.

Combine these two equations to get:

m \cdot a = (\text{Net Force}) = F(\text{tension}) - W.

Therefore:

\begin{aligned}& F(\text{tension})\\ &= m \cdot a + W \\ &= m \cdot (a + g)\\ &= 76\; \rm kg \times \left(1.3\; \rm m \cdot s^{-2} + 9.8\; \rm m \cdot s^{-2}\right)\\ &\approx 8.4\times 10^{2}\; \rm N\end{aligned}.

By Newton's Third Law of motion, Spiderman would exert a force of the same size on the strand of web. Hence, the size of the force in the strand of the web should be approximately 8.4\times 10^{2}\; \rm N (downwards.)

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A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th
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1. 0.42 Hz

The frequency of a simple harmonic motion for a spring is given by:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

where

k = 7 N/m is the spring constant

m = 1 kg is the mass attached to the spring

Substituting these numbers into the formula, we find

f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz

2. 2.38 s

The period of the harmonic motion is equal to the reciprocal of the frequency:

T=\frac{1}{f}

where f = 0.42 Hz is the frequency. Substituting into the formula, we find

T=\frac{1}{0.42 Hz}=2.38 s

3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: x=0, so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:

E=K=\frac{1}{2}mv^2

where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

E=U=\frac{1}{2}kA^2

Since the total energy must be conserved, we have:

\frac{1}{2}mv^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{m}{k}}v=\sqrt{\frac{1 kg}{7 N/m}}(0.5 m/s)=0.19 m

5. Amplitude of the motion: 0.44 m

We can use again the law of conservation of energy.

- E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2 is the initial mechanical energy of the system, with x_0=0.4 m being the initial displacement of the mass and v_0=0.5 m/s being the initial velocity

- E_f = \frac{1}{2}kA^2 is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m

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We can use again the law of conservation of energy.

- E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2 is the initial mechanical energy of the system, with x_0=0.4 m being the initial displacement of the mass and v_0=0.5 m/s being the initial velocity

- E_f = \frac{1}{2}mv_{max}^2 is the mechanical energy of the system when x=0, which is when the system has maximum velocity, v_{max}

Equalizing the two expressions, we can solve to find v_{max}, the maximum velocity:

\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m

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