{"id":3325,"date":"2018-07-08T11:08:50","date_gmt":"2018-07-08T18:08:50","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=3325"},"modified":"2018-07-08T11:08:50","modified_gmt":"2018-07-08T18:08:50","slug":"simple-harmonic-motion-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/simple-harmonic-motion-problems-and-solutions.htm","title":{"rendered":"Simple harmonic motion – problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. An object vibrates with a frequency of 5 Hz to rightward and leftward. The object moves from equilibrium point to the maximum displacement<\/a> at rightward. Determine the time interval required to reach the maximum displacement at rightward eleven times.<\/span><\/span><\/p>\n

A. 2.05<\/span><\/span><\/p>\n

B. 2.20<\/span><\/span><\/p>\n

C. 2.25<\/span><\/span><\/p>\n

D. 2.50<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Frequency (f) = <\/span><\/span>the amount of vibration for 1 second<\/span><\/span><\/span> = 5 Hz<\/span><\/span><\/p>\n

Period (T) = the <\/span><\/span>time interval to do one vibration <\/span><\/span><\/span>= 1\/f = 1\/5 = 0.2 seconds<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> The time interval required to reach the maximum displacement at rightward eleven times<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

\"SimpleThe pattern of the object vibration :<\/span><\/span><\/span><\/p>\n

(1 vibration) : B \u2192 C \u2192 B \u2192 A \u2192 B <\/span><\/span><\/p>\n

For one vibration, the object performs four vibrations that are B to C, C to B, B to A, A to B. The time interval required for a single vibration is 0.2 seconds \/ 4 = 0.05 seconds.<\/span><\/span><\/span><\/p>\n

The time interval required to reach the maximum displacement at rightward eleven times <\/span><\/span><\/span>= (10 x 0.2 seconds) + 0.05 seconds = 2 seconds + 0.05 seconds = 2.05 seconds.<\/span><\/span><\/span><\/p>\n

The correct answer is A.<\/span><\/span><\/p>\n

2. <\/span><\/span>A spring is hung with an object and vibrated. For the vibration frequency to double the original vibration frequency, then the mass of the object is changed to\u2026<\/span><\/span><\/span><\/p>\n

A. twice the mass of the original load<\/span><\/span><\/span><\/p>\n

B. four times the mass of the original load<\/span><\/span><\/span><\/p>\n

C. half the load mass time<\/span><\/span><\/span><\/p>\n

D. a quarter of the original load mass<\/span><\/span><\/span><\/p>\n

Solution :<\/span><\/span><\/p>\n

The equation of the frequency of the spring’s vibration :<\/span><\/span><\/p>\n

\"Simple<\/p>\n

f = frequency, k = constant, m = mass of object<\/i><\/span><\/span><\/p>\n

Frequency (f) of spring’s vibration if k = 1 time, m = 1 time :<\/span><\/span><\/p>\n

\"Simple<\/p>\n

For the frequency of (f) the spring vibration to be twice the mass (m) is changed to 0.25 times or 1\/4 times:<\/span><\/span><\/span><\/p>\n

\"Simple<\/p>\n

The correct answer is D. <\/span><\/span><\/p>\n

3. <\/span><\/span>A spring with a constant k = 1000 N \/ m is hung with an object with a mass of 400 grams. The object is pulled to the right as far as 5 cm, then released, so the object is simple oscillating harmonics<\/a>. Determine the amplitude and frequency of the object oscillation.<\/span><\/span><\/span><\/p>\n

\"Simple<\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

Spring’s constant (k) = 1000 N\/m<\/span><\/span><\/p>\n

Mass<\/a> of object (m) = 400 gram = 0.4 kg<\/span><\/span><\/p>\n

Amplitude (\u0394x) = 5 cm = 0.05 m<\/span><\/span><\/p>\n

Wanted :<\/u><\/span><\/span> Amplitude and frequency of oscillation<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

The amplitude of spring oscillation = \u0394x = 5 cm.<\/span><\/span><\/p>\n

The frequency of oscillation :<\/span><\/span><\/p>\n

\"Simple<\/p>\n

The correct answer is B.<\/span><\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

1. An object vibrates with a frequency of 5 Hz to rightward and leftward. The object moves from equilibrium point to the maximum displacement at rightward. Determine the time interval required to reach the maximum displacement at rightward eleven times. A. 2.05 B. 2.20 C. 2.25 D. 2.50 Known :<\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"","_seopress_robots_index":"","_seopress_robots_follow":"","_seopress_robots_imageindex":"","_seopress_robots_snippet":"","_seopress_robots_primary_cat":"","_seopress_robots_breadcrumbs":"","_seopress_robots_freeze_modified_date":"","_seopress_robots_custom_modified_date":"","_seopress_robots_canonical":"","_seopress_social_fb_title":"","_seopress_social_fb_desc":"","_seopress_social_fb_img":"","_seopress_social_fb_img_attachment_id":0,"_seopress_social_fb_img_width":0,"_seopress_social_fb_img_height":0,"_seopress_social_twitter_title":"","_seopress_social_twitter_desc":"","_seopress_social_twitter_img":"","_seopress_social_twitter_img_attachment_id":0,"_seopress_social_twitter_img_width":0,"_seopress_social_twitter_img_height":0,"_seopress_redirections_value":"","_seopress_redirections_enabled":"","_seopress_redirections_enabled_regex":"","_seopress_redirections_logged_status":"","_seopress_redirections_param":"","_seopress_redirections_type":0,"_seopress_analysis_target_kw":"Simple harmonic motion - problems and solutions","_seopress_news_disabled":"","_seopress_video_disabled":"","_seopress_video":[],"_seopress_pro_schemas_manual":[],"_seopress_pro_rich_snippets_disable_all":"","_seopress_pro_rich_snippets_disable":[],"_seopress_pro_schemas":[],"footnotes":""},"categories":[3],"tags":[],"class_list":["post-3325","post","type-post","status-publish","format-standard","hentry","category-solved-problems-in-basic-physics"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/3325","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/comments?post=3325"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/3325\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=3325"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=3325"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=3325"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}