Are the 2nd and 3rd problems in this video correctly solved? (charged dipole and organ pipe problems)

[WMDhamnekar]

Homework Statement

Q.2:The two charges of $-4\mu C$ and $+ 4\mu C$are placed at the point A(1,0.4)m and B( 2,-1,5)m located in an electric field $\overrightarrow{E}=0.20 \hat{i} V/cm.$The magnitude of Torque acting on the dipole is $8\sqrt{\alpha} \times 10^{-5} Nm$ What is the value of $\alpha?$
Q.3 :A closed organ pipe 150 cm long gives 7 beats per second with an open organ pipe of length 350 cm, both vibrating in fundamental mode. What is the velocity of sound in meter per second?

Relevant Equations

Not applicable

Relevant video $\Rightarrow$ Physics questions and answers

My answer to Q.2

Let's break down the problem and solve it step-by-step:​

1. Find the dipole moment (p):

• The dipole moment is defined as the product of the charge (q) and the separation distance (d) between the charges.
• p = qd
• Here, $q = 4 \mu C = 4 × 10^{-6} C$
• To find the separation distance, we need to find the vector difference between the positions of the two charges:$d = B - A = (2 - 1)\hat{i} + (-1 - 0)\hat{j} + (5 - 4)k\hat{k} = \hat{i} - \hat{j} + \hat{k}$
• The magnitude of d is $|d| = \sqrt(1^2 + (-1)^2 + 1^2) = \sqrt{3} m$
• So, the dipole moment $\rho = (4 × 10^{-6} C) \times (\sqrt{3} m) = 4\sqrt{3} \times 10^{-6} C⋅m$

2. Find the torque $(\tau):$

• The torque on a dipole in a uniform electric field is given by:$\tau = \rho \times E$
• Here, $E = 0.20 \hat{i} V/cm = 20 \hat{i} V/m$
• So, $\tau = (4\sqrt{3} \times 10^{-6} C⋅m) \times (20 \hat{i} V/m)$
• Calculating the cross product, we get:$\tau = 80\sqrt{3} \times 10^{-6} (\hat{j} + \hat{k}) N⋅m$

3. Find the magnitude of the torque: $|\tau| = 80\sqrt{3} \times 10^{-6} \sqrt{(1^2 + 1^2)} N \cdot m = 80\sqrt{6} \times 10^{-6} N \cdot m$
4. Compare with the given magnitude:We are given that the magnitude of the torque is $8\sqrt{\alpha} \times 10^{-5} N⋅m.$So, $80\sqrt{}6 \times 10^{-6} = 8\sqrt{\alpha} \times 10^{-5}$
Simplifying, we get:$\sqrt{6} = \sqrt{\alpha}$
Therefore, $\alpha =6$ But author of video said $\alpha =2$
My answer to Q.3
Let's denote the frequency of the closed organ pipe as $f_1$ and the frequency of the open organ pipe as $f_2.$
Fundamental Frequency of a Closed Organ Pipe: $f_1 = v / 4L_1$where:

• $f_1$ is the fundamental frequency
• v is the velocity of sound
• $L_1$ is the length of the closed pipe

Fundamental Frequency of an Open Organ Pipe: $f_2 = v / 2L_2$where:

• $f_2$ is the fundamental frequency
• v is the velocity of sound
• $L_2$ is the length of the open pipe

Given:

• $L_1 = 150 cm = 1.5 m$
• $L_2 = 350 cm = 3.5 m$
• Beat frequency = 7 Hz (This means $|f_1 - f_2| = 7)$

Let's assume f₁ > f₂: $f_1 - f_2 = 7$
Substituting the expressions for $f_1$ and $f_2:$
$(v / 4L_1) - (v / 2L_2) = 7$
$v(1/4L_1 - 1/2L_2) = 7$
$v[(2L_2 - 4L_1) / (4L_1L_2)] = 7$
$v = 7 * (4L_1L_2) / (2L_2 - 4L_1)$
Substituting the given values:
v = 7 * (4 * 1.5 * 3.5) / (2 * 3.5 - 4 * 1.5)
v = 7 * 21 / 1
v = 147 m/s
Therefore, the velocity of sound in air is 147 m/s. But the author said velocity of sound is 294 m/s. How is that?

 

[kuruman]

In the dipole problem your calculation of the dipole moment vector is incorrect. Use the definition $$\mathbf p=\sum_i q_i\mathbf r_i.$$ Here, let the positive charge have subscript 2 and the negative charge subscript 1. Then $$\mathbf p= q~\mathbf r_2-q~\mathbf r_1=q\{x_2-x_1,y_2-y_1,z_2-z_1 \}.$$The definition gives you the vector directly which you can then cross with the electric field. You don't need to mess with magnitudes.

 

[kuruman]

In the beats problem do you think that it is an accident that your answer is exactly half of the given answer? How is the difference of frequencies related to the beat frequency? Look it up.

For future reference, if you you have two separate problems, please post them on separate threads. You don't want to provoke the displeasure of the mentors.

 

[WMDhamnekar]

There was a computation error in my answer to Q.3.
$\frac{V}{4L_1} -\frac{V}{2L_2}= 7 \Rightarrow V\big(\frac{1}{4L_1}- \frac{1}{2L_2}\big)= 7 \Rightarrow \frac{V(L_2-2L_1)}{4L_1L_2} =7 \Rightarrow \frac{7*4L_1L_2}{L_2- 2L_1}= \frac{7*4*1.5*3.5}{3.5-(2*1.5)}= 294 m/s$

Hence Author of the video computed the answers for 2nd and 3rd problem correctly.