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Question: A string \[2.0m\] long and fixed at its ends is driven by a 240 Hz vibrator. The string vibrates in ...

A string 2.0m2.0m long and fixed at its ends is driven by a 240 Hz vibrator. The string vibrates in its third harmonic mode. The speed of the wave and its fundamental frequency are?
(A) 320 m/s, 120 Hz
(B) 180 m/s, 80 Hz
(C) 180 m/s, 120 Hz
(D) 320 m/s, 80 Hz

Explanation

Solution

A string fixed at its ends signifies that it has nodes at its ends. The third harmonic mode has a frequency which is simply thrice the frequency of the fundamental mode.
Formula used: In this solution we will be using the following formulae;
f0=v2l{f_0} = \dfrac{v}{{2l}} where f0{f_0} is the fundamental frequency of string fixed at both ends. vv is the speed of the wave on the string, and ll is the length of the string.
f2=3f0{f_2} = 3{f_0} where f2{f_2} is the third harmonic frequency (also known as the second overtone frequency)

Complete Step-by-Step Solution:
A particular vibrator at a particular frequency is said to be vibrating a string of a particular length and fixed at its ends. This string is said to vibrate at its third harmonic mode (also commonly called second overtone mode). We are to determine the speed of the wave and the fundamental frequency.
To calculate the fundamental frequency, we shall note that the third harmonics can be given as
f2=3f0{f_2} = 3{f_0} where f2{f_2} is the third harmonic frequency (also known as the second overtone frequency) and f0{f_0} is the fundamental frequency, hence,
f0=f23{f_0} = \dfrac{{{f_2}}}{3} which by inserting given values, we get,
f0=2403=80Hz{f_0} = \dfrac{{240}}{3} = 80Hz
Now to calculate the velocity, we recall that the fundamental frequency can be given as
f0=v2l{f_0} = \dfrac{v}{{2l}} where vv is the speed of the wave on the string, and ll is the length of the string.
Hence, we get
v=2lf0=2×2×80v = 2l{f_0} = 2 \times 2 \times 80
v=320m/s\Rightarrow v = 320m/s

Hence, the correct option is D

Note: Alternatively, we could calculate first, the speed of light from
f2=3f0=3v2l{f_2} = 3{f_0} = 3\dfrac{v}{{2l}}
Hence, v=2lf23v = \dfrac{{2l{f_2}}}{3} which by inserting values, we get,
v=2×2×2403=320m/sv = \dfrac{{2 \times 2 \times 240}}{3} = 320m/s
And then the fundamental frequency can be calculated as
f2=v2l=3202×2=80Hz{f_2} = \dfrac{v}{{2l}} = \dfrac{{320}}{{2 \times 2}} = 80Hz