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3 years ago

11

To what tension (in newtons) must you adjust the screw so that a transverse wave of wavelength 3.33 cm makes 621 vibrations per

second?

1 answer:

PilotLPTM [1.2K]3 years ago

7 0

Answer:

T=9.4 N

Explanation:

We are given that

Mass of wire,m=16.5 g= $\frac{16.5}{1000}$ kg

1 kg=1000g

Length of wire,l=75 cm= $75\times 10^{-2}$ m

1 m=100 cm

Wavelength of transverse wave= $\lambda=3.33 cm=3.33\times 10^{-2}$ m

Frequency= $621 Hz$

Mass per unit length= $m_l=\frac{m}{l}=\frac{16.5}{1000\times 75\times 10^{-2}}=0.022 kg/m$

$\nu=\frac{v}{\lambda}$

$v=\nu \lambda$

Where $\nu=$ frequency of wave

$\lambda$ =Wavelength of wave

Speed of wave=v

Using the formula

$v=3.33\times 10^{-2}\times 621=20.7m/s$

$v=\sqrt{\frac{T}{m_l}}$

$v^2=\frac{T}{m_l}$

$T=v^2m_l$

Using the formula

$T=(20.7)^2\times 0.022=9.4N$

Hence, the tension,T=9.4 N

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Answer:

8.6 m/s

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For the dog in the problem, we have

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A cylinder of radius R, length L, and mass M is released from rest on a slope inclined at angle θ. It is oriented to roll straig

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Answer:

$\mu_s=\frac{1}{3}\tan \theta$

Explanation:

Let the minimum coefficient of static friction be $\mu_s$ .

Given:

Mass of the cylinder = $M$

Radius of the cylinder = $R$

Length of the cylinder = $L$

Angle of inclination = $\theta$

Initial velocity of the cylinder (Released from rest) = 0

Since, the cylinder is translating and rolling down the incline, it has both translational and rotational motion. So, we need to consider the effect of moment of Inertia also.

We know that, for a rolling object, torque acting on it is given as the product of moment of inertia and its angular acceleration. So,

$\tau =I\alpha$

Now, angular acceleration is given as:

$\alpha = \frac{a}{R}\\Where, a\rightarrow \textrm{linear acceleration of the cylinder}$

Also, moment of inertia for a cylinder is given as:

$I=\frac{MR^2}{2}$

Therefore, the torque acting on the cylinder can be rewritten as:

$\tau = \frac{MR^2}{2}\times \frac{a}{R}=\frac{MRa}{2}------ 1$

Consider the free body diagram of the cylinder on the incline. The forces acting along the incline are $mg\sin \theta\ and\ f$ . The net force acting along the incline is given as:

$F_{net}=Mg\sin \theta-f\\But,\ f=\mu_s N\\So, F_{net}=Mg\sin \theta -\mu_s N-------- 2$

Now, consider the forces acting perpendicular to the incline. As there is no motion in the perpendicular direction, net force is zero.

So, $N=Mg\cos \theta$

Plugging in $N=Mg\cos \theta$ in equation (2), we get

$F_{net}=Mg\sin \theta -\mu_s Mg\cos \theta\\F_{net}=Mg(\sin \theta-\mu_s \cos \theta)--------------3$

Now, as per Newton's second law,

$F_{net}=Ma\\Mg(\sin \theta-\mu_s \cos \theta)=Ma\\\therefore a=g(\sin \theta-\mu_s \cos \theta)------4$

Now, torque acting on the cylinder is provided by the frictional force and is given as the product of frictional force and radius of the cylinder.

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Therefore, the minimum coefficient of static friction needed for the cylinder to roll down without slipping is given as:

$\mu_s=\frac{1}{3}\tan \theta$

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