佳礼资讯网

 找回密码
 注册

ADVERTISEMENT

搜索
楼主: 辉文

物理讨论专区

   关闭 [复制链接]
发表于 3-9-2010 01:20 PM | 显示全部楼层
回复 440# Winterblade
amplitude= (  f1² - f2² )/8
simplify, u obtain  y=  ( f1² - f2² )/8 . sin  4(pi)t
回复

使用道具 举报


ADVERTISEMENT

发表于 3-9-2010 01:54 PM | 显示全部楼层
so the stationary wave equation is actually for the particlar wave de?? and if we wan to calculate for the displacement of particular element at particular distance, we sub nto cos kx?? how bout the time??
回复

使用道具 举报

发表于 3-9-2010 01:55 PM | 显示全部楼层
2cos x = amplitude then y is what??
回复

使用道具 举报

发表于 3-9-2010 09:12 PM | 显示全部楼层
yes stationary wave equation
when find color=Red]max  displacement, we can sub into the cos kx
when we want to find a particular displacement of a element at a particular distance at time t, we just insert all the info into the equation.  
2cos x = amplitude then y is what??
that it is just an example so  dun bother it but
if this is wave equation, then  y is displacement of the particle at time t
回复

使用道具 举报

发表于 4-9-2010 07:51 PM | 显示全部楼层
我有一题physic 问题, 请各位大大帮帮忙。。
The distance of the Earth from the Sun is 1.5 x 10^11 m. The mass of the Sun is (3.24 x 10^5)M where M is the mass of the Earth.
Find the distance from the Earth along the line joining the Earth to the Sun where the gravitational attraction of the Sun is equal that of the Earth.
回复

使用道具 举报

发表于 4-9-2010 09:26 PM | 显示全部楼层
回复 445# jennykuwn
assume a body mass m is situated between earth and sun
distance between earth and mass=r
distance between sun and mass = 1.5 x 10^11 - r
  force earth = force sun
use F=GMm/r^2
GMm/r^2 = G(3.24 x 10^5 M)m / (1.5 x 10^11 - r)^2
1/r^2 = (3.24 x 10^5 )/ (1.5 x 10^11 - r)^2
solve, using quadratic formula,  u will get the r.
r=2.63 x 10^8 m
回复

使用道具 举报

Follow Us
发表于 5-9-2010 05:38 PM | 显示全部楼层
I still cant differentiate between the two..y= cos kx sin wt, the y is calculate for what and the cos kx is what?
回复

使用道具 举报

发表于 5-9-2010 09:51 PM | 显示全部楼层
Two sound waves with frequencies 450Hz and 458Hz undergo superposition. Calculate the
i) frequency of the resultant wave
ii) beat

how to calculate ar???haiz..dun understand..dun quite understand actually about beat
回复

使用道具 举报


ADVERTISEMENT

发表于 6-9-2010 05:18 PM | 显示全部楼层
1.y is displacement of particle at time t, cos kx is amplitude ,  always remember amplitude depends on the fixed x.

for Q 2,
resultant wave frequency = (f1+f2)/2
beat frequency=f1-f2
beat frequency mean no of beats u heard in 1 s time
resultant frequency is the no of complete oscillations made by the two waves in 1 s
回复

使用道具 举报

发表于 6-9-2010 09:26 PM | 显示全部楼层
then calculate maximum displacement why you use cos kx?? isnt that y= ___?
回复

使用道具 举报

发表于 6-9-2010 09:29 PM | 显示全部楼层
amplitude is maximum displacement..it does not depends on t meh?? the y does not depends on x?
回复

使用道具 举报

发表于 7-9-2010 02:52 PM | 显示全部楼层
yes, calculate displacement we use y =cos kx sin wt
but y is MAX displacement  when  at sin wt =1
so, max displacement= cos kx
y is depends on t ,amplitude depends on  x , x is fixed
回复

使用道具 举报

发表于 7-9-2010 06:06 PM | 显示全部楼层
Can give me some others example questions??
回复

使用道具 举报

发表于 7-9-2010 09:27 PM | 显示全部楼层
y calculate the displacement of what?? particle on the wave? or the wave?
回复

使用道具 举报

发表于 10-9-2010 11:07 AM | 显示全部楼层
One ques here

A transverse sinusoidal wave on a string has a period T= 25.0 ms and travels in the negative x direction with a speed of 30.0 ms -1. At t=0 s, a particle on the string at x = 0m has a trasverse position of 2.0m and is travelling downwards with a speed of 2.0 ms-1. Calculate the amplitude, maximum speed of particle and wavelength..

I dun understand why they give the particle's velocity..use for what?
回复

使用道具 举报

发表于 10-9-2010 11:09 AM | 显示全部楼层
One ques here

A transverse sinusoidal wave on a string has a period T= 25.0 ms and travels in the negative x direction with a speed of 30.0 ms -1. At t=0 s, a particle on the string at x = 0m has a trasverse position of 2.0m and is travelling downwards with a speed of 2.0 ms-1. Calculate the amplitude, maximum speed of particle and wavelength..

I dun understand why they give the particle's velocity..use for what?
回复

使用道具 举报


ADVERTISEMENT

发表于 11-9-2010 01:28 PM | 显示全部楼层
回复 456# vick5821


    You got the answers?
回复

使用道具 举报

发表于 11-9-2010 09:10 PM | 显示全部楼层
Yes...我有答案
回复

使用道具 举报

发表于 11-9-2010 10:25 PM | 显示全部楼层
At what point on a stationary waves that maximum sound is heard, the node or antinode?
回复

使用道具 举报

发表于 11-9-2010 10:32 PM | 显示全部楼层
回复 458# vick5821


    oo please give me
回复

使用道具 举报

您需要登录后才可以回帖 登录 | 注册

本版积分规则

 

所属分类: 欢乐校园


ADVERTISEMENT



ADVERTISEMENT



ADVERTISEMENT

ADVERTISEMENT


版权所有 © 1996-2026 Cari Internet Sdn Bhd (483575-W)|IPSERVERONE 提供云主机|广告刊登|关于我们|私隐权|免控|投诉|联络|脸书|佳礼资讯网

GMT+8, 31-3-2026 10:22 PM , Processed in 0.088783 second(s), 19 queries , Gzip On.

Powered by Discuz! X3.4

Copyright © 2001-2021, Tencent Cloud.

快速回复 返回顶部 返回列表