The figure below shows pressure variation in | vedclass

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(D) The correct option is $(d)$.From the given figures, we can observe the following:$1$. Frequency: Wave $1$ completes more oscillations in the same

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Wave $1$ has higher frequency and smaller amplitude compared to wave $2$.

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(D) The correct option is $(d)$.From the given figures, we can observe the following:$1$. Frequency: Wave $1$ completes more oscillations in the same time interval compared to wave $2$. Since frequency $f = \frac{1}{T}$, where $T$ is the time period, a higher number of oscillations implies a higher frequency. Thus, wave $1$ has a higher frequency than wave $2$.$2$. Amplitude: The amplitude is represented by the maximum displacement from the mean position (pressure variation). Visually, the peak height of wave $1$ is smaller than the peak height of wave $2$. Therefore, wave $1$ has a smaller amplitude compared to wave $2$.Note: The original option $(d)$ stated "shorter wavelength". In a time-domain graph, the horizontal axis represents time, so the distance between peaks corresponds to the time period $T$. A higher frequency corresponds to a shorter time period. Since wavelength $\lambda = vT$ (where $v$ is the speed of sound), a shorter time period implies a shorter wavelength. Thus, wave $1$ has a shorter wavelength and smaller amplitude compared to wave $2$.

The figure below shows pressure variation in two different sound waves in air with time at a given position. Both the figures are drawn to the same scale. Which of the following statements is true?

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