GRE Physics Companion: Waves on Strings: Reflection, Transmission, and Mechanical Impedance

GRE Physics Companion: Waves on Strings: Reflection, Transmission, and Mechanical Impedance

For an ideal string,

(1)

At a lossless junction,

(2)

The power fractions are

(3)

1 High-value GRE facts

1. Z = T∕c = μc = .
2. Displacement continuity gives 1 + r = t.
3. force continuity gives Z₁(1 − r) = Z₂t.
4. r = (Z₁ − Z₂)∕(Z₁ + Z₂).
5. t = 2Z₁∕(Z₁ + Z₂).
6. ℛ = |r|².
7. 𝒯 = (Z₂∕Z₁)|t|².
8. A fixed end has r = −1.
9. A free end has r = +1.
10. impedance matching gives r = 0.

Part I: Original GRE-style problems

Problem 1: string impedance

For an ideal string,

1. Z = Tc
2. Z = T∕c
3. Z = c∕T
4. Z = μ∕T
5. Z = Tμc

Problem 2: displacement continuity

At an ideal attached junction,

1. A_i − A_r = A_t
2. A_i + A_r = A_t
3. A_i = A_r + A_t only for equal impedance
4. A_i = A_t
5. A_r = 0 always

Problem 3: reflection amplitude

The displacement reflection coefficient is

1. (Z₂ − Z₁)∕(Z₁ + Z₂)
2. (Z₁ − Z₂)∕(Z₁ + Z₂)
3. 2Z₁∕(Z₁ + Z₂)
4. Z₂∕Z₁
5. Z₁Z₂

Problem 4: transmission amplitude

The displacement transmission coefficient is

1. 2Z₁∕(Z₁ + Z₂)
2. 2Z₂∕(Z₁ + Z₂)
3. (Z₁ − Z₂)∕(Z₁ + Z₂)
4. Z₁∕Z₂
5. always 1

Problem 5: low-to-high impedance reflection

If

(4)

the reflected displacement is

1. inverted
2. noninverted
3. always zero
4. twice the incident amplitude
5. undefined

Problem 6: fixed end

A perfectly fixed end has

1. r = +1
2. r = −1
3. r = 0
4. r = 1∕2
5. r = −1∕2

Problem 7: free end

A perfectly free end has

1. r = +1
2. r = −1
3. r = 0
4. r = 1∕2
5. r = −1∕2

Problem 8: power transmission

The transmitted power fraction is

1. |t|²
2. (Z₁∕Z₂)|t|²
3. (Z₂∕Z₁)|t|²
4. |r|²
5. 1 + |r|²

Problem 9: matched strings

If

(5)

then

1. ℛ = 1
2. r = −1
3. r = 0
4. t = 0
5. no power can cross the interface

Problem 10: numerical reflection

If

(6)

then

(7)

is

1. −1∕3
2. +1∕3
3. 2∕3
4. −2∕3
5. 1

Problem 11: same interface power

For the interface in Problem 10, the transmitted power fraction is

1. 1∕9
2. 2∕9
3. 4∕9
4. 8∕9
5. 1

Problem 12: fixed-free resonance

A string fixed at one end and free at the other satisfies

1. k_n = nπ∕L
2. k_n = (2n − 1)π∕(2L)
3. k_n = 2nπ∕L
4. k_n = nπ∕(2L) for every integer n
5. no standing-wave condition

Part II: Complete worked solutions

Solution 1

For an ideal string,

(8)

Answer: (B).

Solution 2

The attached junction has one displacement:

(9)

Answer: (B).

Solution 3

Combining displacement and force continuity gives

(10)

Answer: (B).

Solution 4

Since

(11)

(12)

Answer: (A).

Solution 5

If

(13)

then

(14)

The reflected displacement is inverted. Answer: (A).

Solution 6

A fixed end requires zero total displacement:

(15)

Thus

(16)

Answer: (B).

Solution 7

A free end requires zero transverse force, which gives

(17)

Thus

(18)

Answer: (A).

Solution 8

Power is proportional to

(19)

Therefore

(20)

Answer: (C).

Solution 9

If

(21)

then

(22)

The interface is impedance matched. Answer: (C).

Solution 10

(23)

Answer: (A).

Solution 11

(24)

Answer: (D).

Solution 12

A fixed reflection contributes phase

(25)

and a free reflection contributes zero displacement phase shift. The round-trip condition gives

(26)

Answer: (B).

2 GRE checklist

1. Compute Z = before deciding the reflection phase.
2. Keep amplitude coefficients separate from power coefficients.
3. Use 1 + r = t at an attached junction.
4. Remember the minus sign from differentiating the reflected spatial phase.
5. Low-to-high impedance gives an inverted reflected displacement.
6. High-to-low impedance gives a noninverted reflected displacement.
7. Fixed end means r = −1; free end means r = +1.
8. Check power conservation with ℛ + 𝒯 = 1.

References

References