In the case of a stationary wave pattern, whi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) $1$. In a stationary wave, the distance between two consecutive nodes or two consecutive antinodes is $\lambda/2$. Thus, option $A$ is incorrect.$

Vedclass · Accepted Answer

In the case of a stretched string, when vibrated, the frequency of the first overtone is the same as the second harmonic.

Vedclass · Answer

(D) $1$. In a stationary wave, the distance between two consecutive nodes or two consecutive antinodes is $\lambda/2$. Thus, option $A$ is incorrect.$2$. In a pipe open at both ends, all harmonics (odd and even) are present. Thus, option $B$ is incorrect.$3$. In a pipe closed at one end, only odd harmonics are present. Thus, option $C$ is incorrect.$4$. For a stretched string, the fundamental frequency is $f_1 = v/2L$. The first overtone is the next higher frequency, which is $f_2 = 2v/2L = 2f_1$. This is also called the second harmonic. Thus, option $D$ is correct.

In the case of a stationary wave pattern, which of the following statements is $CORRECT$?

Similar Questions

$A$ string fixed at both its ends vibrates in $5$ loops as shown in the figure. The total number of nodes and antinodes are respectively:

In a stationary wave,all the particles:

The equation of a stationary wave is $Y = 10 \sin \left( \frac{\pi x}{4} \right) \cos (20 \pi t)$. The distance between two consecutive nodes in metres is

In a standing wave on a string:

Vedclass Test Series

Exam Paper Generator

Online Exam Module