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Waves question

2013 · Shift 1 · Q57
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Waves question

2013 · Shift 1 · Q57

JEE AdvancedPhysicsWavesMultiple correct+4 / −2
A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation y(x, t) = (0.01 m) sin[(62.8 m −-− 1)x] cos[(628 s −-− 1)t] Assuming π\piπ = 3.14, the correct statement(s) is(are)
  1. A
    The number of nodes is 5.
  2. B
    The length of the string is 0.25 m.
  3. C
    The maximum displacement of the mid-point of the string, from its equilibrium position is 0.01 m.
  4. D
    The fundamental frequency is 100 Hz.
View written solutionFree

Correct answer: B, C

  1. Given wave equation

The standing wave on the string is

y(x,t)=(0.01 m)sin⁡[(62.8 m−1)x]cos⁡[(628 s−1)t]y(x,t)=(0.01\,\text{m})\sin[(62.8\,\text{m}^{-1})x]\cos[(628\,\text{s}^{-1})t]y(x,t)=(0.01m)sin[(62.8m−1)x]cos[(628s−1)t]

Comparing with the standard form of a standing wave:

y(x,t)=Asin⁡(kx)cos⁡(ωt)y(x,t)=A\sin(kx)\cos(\omega t)y(x,t)=Asin(kx)cos(ωt)

we get:

  • Amplitude factor: A=0.01 mA=0.01\,\text{m}A=0.01m
  • Wave number: k=62.8 m−1k=62.8\,\text{m}^{-1}k=62.8m−1
  • Angular frequency: ω=628 s−1\omega=628\,\text{s}^{-1}ω=628s−1

The string is vibrating in the fifth harmonic.

  1. Use harmonic condition for a string fixed at both ends

For a string fixed at both ends, in the nnn-th harmonic:

L=nλ2L=n\frac{\lambda}{2}L=n2λ​

Here n=5n=5n=5, so

L=5λ2L=5\frac{\lambda}{2}L=52λ​

Also,

k=2πλ⇒λ=2πkk=\frac{2\pi}{\lambda} \Rightarrow \lambda=\frac{2\pi}{k}k=λ2π​⇒λ=k2π​

Substitute k=62.8k=62.8k=62.8 and π=3.14\pi=3.14π=3.14:

λ=2×3.1462.8=6.2862.8=0.1 m\lambda=\frac{2\times 3.14}{62.8} = \frac{6.28}{62.8}=0.1\,\text{m}λ=62.82×3.14​=62.86.28​=0.1m

Hence,

L=5⋅0.12=5⋅0.05=0.25 mL=5\cdot \frac{0.1}{2}=5\cdot 0.05=0.25\,\text{m}L=5⋅20.1​=5⋅0.05=0.25m

So Option B is correct.

  1. Check number of nodes

For the nnn-th harmonic of a string fixed at both ends, the number of nodes is:

n+1n+1n+1

Since n=5n=5n=5,

number of nodes=5+1=6\text{number of nodes}=5+1=6number of nodes=5+1=6

So the statement “The number of nodes is 5” is false.

Thus Option A is incorrect.

  1. Check displacement of the midpoint

The midpoint of the string is at

x=L2=0.252=0.125 mx=\frac{L}{2}=\frac{0.25}{2}=0.125\,\text{m}x=2L​=20.25​=0.125m

Its displacement is:

y(L2,t)=0.01sin⁡(62.8×0.125)cos⁡(628t)y\left(\frac{L}{2},t\right)=0.01\sin(62.8\times 0.125)\cos(628t)y(2L​,t)=0.01sin(62.8×0.125)cos(628t)

Now,

62.8×0.125=7.8562.8\times 0.125=7.8562.8×0.125=7.85

Since π=3.14\pi=3.14π=3.14,

7.85=5π27.85=\frac{5\pi}{2}7.85=25π​

Therefore,

sin⁡(5π2)=1\sin\left(\frac{5\pi}{2}\right)=1sin(25π​)=1

So,

y(L2,t)=0.01cos⁡(628t)y\left(\frac{L}{2},t\right)=0.01\cos(628t)y(2L​,t)=0.01cos(628t)

Hence the maximum displacement of the midpoint is

0.01 m0.01\,\text{m}0.01m

So Option C is correct.

  1. Check fundamental frequency

Frequency of this harmonic is:

f_5=\frac{\omega}{2\pi}=\frac{628}{2\times 3.14}= rac{628}{6.28}=100\,\text{Hz}

But this is the fifth harmonic frequency, not the fundamental frequency.

So the fundamental frequency is

f1=f55=1005=20 Hzf_1=\frac{f_5}{5}=\frac{100}{5}=20\,\text{Hz}f1​=5f5​​=5100​=20Hz

Therefore the statement “The fundamental frequency is 100 Hz” is false.

So Option D is incorrect.

  1. Final conclusion

Correct statements are:

  • B: The length of the string is 0.25 m0.25\,\text{m}0.25m.
  • C: The maximum displacement of the midpoint is 0.01 m0.01\,\text{m}0.01m.

Thus the correct answer is B, C.

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