A sound wave of frequency `440 Hz` is passing through in air. An `O_(2)` molecule `( mass = 5.3 xx 10^(- 26) kg)` is set in oscillation with an amplitude of `10 ^(-6) m`. Its speed at the centre of its oscillation isA. `1.70 xx 10^(-5) m//s`B. `17.0 xx 10^(-5) m//s`C. `2.76 xx 10^(-3) m//s`D. `2.77 xx 10^(-5) m//s`

A sound wave of frequency `440 Hz` is passing through in air. An `O_(2

1.	A sound wave of frequency `440 Hz` is passing through in air. An `O_(2)` molecule `( mass = 5.3 xx 10^(- 26) kg)` is set in oscillation with an amplitude of `10 ^(-6) m`. Its speed at the centre of its oscillation isA. `1.70 xx 10^(-5) m//s`B. `17.0 xx 10^(-5) m//s`C. `2.76 xx 10^(-3) m//s`D. `2.77 xx 10^(-5) m//s`
Answer» Correct Answer - C `v_(max) = omega_(n) A = ( 2 pi f) A = ( 2 pi) = (2 pi) (440) (10^(-6))` ` = 2.76 xx 10^(-3) m//s`

Discussion

No Comment Found

Related InterviewSolutions

The equation of stationary wave along a stretched string is given by `y=5 sin ((pix)/(3)) cos 40 pi t `, where x and y are in cm and t in second. The separation between two adjacent nodes is
In stationary waves, distance between a node and its nearest antinode is 20 cm . The phase difference between two particles having a separation of 60 cm will be
The equation of a stationary wave is ` y = 0.8 cos ((pi x)/(20)) sin 200 pi t` where ` x` is in cm and t is in s. The separation between consecutive nodes will beA. `20 cm`B. `10 cm`C. `40 cm`D. `30 cm`
Two waves are approaching each other with a velocity of `20m//s` and frequency `n`. The distance between two consecutive nudes isA. `(20)/(n)`B. `(10)/(n)`C. `(5)/(n)`D. `(n)/(10)`
Spacing between two successive nodes in a standing wave on a string is `x`. If frequency of the standing wave is kept unchanged but tension in the string is doubled, then new sapcing between successive nodes will become:A. `x//sqrt(2)`B. `sqrt(2)x`C. `x//2`D. 2x
If `v_(1) , v_(2) and v_(3)` are the fundamental frequencies of three segments of stretched string , then the fundamental frequency of the overall string isA. `v_(1) + v_(2) + v_(3)`B. `[(1)/(v_(1)) + (1)/(v_(2)) + (1)/( v_(3))]^(-1)`C. `v_(1) v_(2) v_(3)`D. `[v_(1) v_(2) v_(3)]^(1//3)`
If the length of a stretched string is shortened by `40 %` and the tension is increased by `44 %`, then the ratio of the final and initial fundamental frequencies isA. `3 : 4`B. `4 : 3`C. `1 : 3`D. `2 :1`
To decrease the fundamental frequency of a stretched string fixed at both ends one mightA. increase its tensionB. increase its wave velocityC. increase its lengthD. decrease its linear mass density
A stretched string of length ` 1 m` fixed at both ends , having a mass of `5 xx 10^(-4) kg` is under a tension of `20 N`. It is plucked at a point situated at `25 cm` from one end . The stretched string would vibrate with a frequency ofA. `400 Hz`B. `100 Hz`C. `200 Hz`D. `256 Hz`
A sonometer wire , `100 cm` in length has fundamental frequency of `330 Hz`. The velocity of propagation of tranverse waves along the wire isA. `330 m//s`B. `660 m//s`C. `115 m//s`D. `990 m//s`

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment